Ciass _QA_33_ Book_JS_g__ Copyright i\° CDRrfRIGHT DEPOSm FUNDAMENTALS )F HIGH SCHOOL MATHEMATICS RUG G -CLARK I RUGG-CLARK MATHEMATICS TEXTS By Harold O. Rugg and John R. Clark Fundamentals of High School Mathematics A Textbook Designed to Follow Arithmetic BY HAROLD O. RUGG Associate Professor of Education University of Chicago AND JOHN R. CLARK Head of Department of Mathematics Francis W. Parker School, Chicago Experimental Edition, distributed at cost for experimental purposes only. NOT for general commercial distribution 1918 WORLD BOOK COMPANY YONKERS-ON-HUDSON, NEW YORK 2126 PRAIRIE AVENUE, CHICAGO WORLD BOOK COMPANY THE HOUSE OF APPLIED KNOWLEDGE Established, 1905, by Caspar W. Hodgson yonkers-on-hudson, new york 2126 Prairie Avenue, Chicago The purpose of this house is to publish books that apply the world's knowledge to the world's needs. Public-school texts in mathematics should be designed in terms of careful studies of social needs as well as in terms of methods by which pupils learn. This book on general mathematics, the first of the Rugg-Clark Mathematics Texts, is the outgrowth of years of scientific investigation and of experimental teaching. It will be followed by texts for Junior High Schools and for Elementary Schools, each of which will be published only after years of scien- tific investigation and experimentation. SEP -3 1918 ©CI.A503268 rcmt: fhsm-i Copyright, 1918, by World Book Company All rights reserved u.^/lr^ OJMMJ& QmulaJUm^j FOREWORD THE CONSTRUCTION OF COURSES OF STUDY IN HIGH SCHOOL MATHEMATICS The one-year course of study represented by this textbook is the outcome of five years of critical investigation of high-school mathematics. A recent monograph published by the writers * presents in detail the evidence developed by these years of in- vestigation. It reveals striking weaknesses in the content and teaching of first-year algebra. It shows that the achievement of more than 22.000 pupils on carefully designed standardized tests is very unsatisfactory ; that the present course of study is not organized primarily to provide an opportunity for training in problem-solving ; and that much of the material included in the traditional course will never be used in actual life situations. The evidence is clear, therefore, that the present course of study in first-year algebra must be completely reconstructed. The study of the historical development of the present first- year high school algebra shows that it has come down from its former position in the upper college curriculum with but two slight modifications of content and presentation. One has been revealed in the increased emphasis on the formal and manipu- lative aspects : the second in the slight improvements which have been made in methods of presentation through the use of more concrete devices. On the whole, however, the course has retained that emphasis upon rigorous, logical organization which characterizes the mathematical thinking of " mathematicians." The need is evident, therefore, for a course of study which will avoid the two fundamental weaknesses pointed out above ; namely, first, a course which will eliminate formal non-essentials, basing such elimination on scientific investigation ; second, one which will tie together, in a psychologically and sequentially worked-out scheme, the fundamental mathematical notions and tool operations which are needed by adults to facilitate quanti- 1 Scientific Method in the Reco?istmction of Xinth- Grade Mathematics. University of Chicago Press. 1918. iii iv Foreword tative thinking. The course of study presented in this text is therefore unique in these particulars. PRINCIPLES OF TEXTBOOK DESIGN Three important criteria have been used to determine the content and the organization of this experimental course of study : first, subject matter must be organized in terms of a real psychological analysis of " learning " in mathematics ; second, those mathematical notions and devices which have the widest application must be emphasized roughly in proportion to their frequency of use ; thus the material has been selected in terms of the social criterion ; third, the course must be organized to provide the pupil with the maximum opportunity to do genuine thinking, real problem-solving, rather than to emphasize the drill or manipulative aspects which now commonly require most of the pupil's time. The first criterion. To organize a course of study in terms of a real psychology of " learning" in mathematics shows us that new meanings, new concepts which are to be learned, must be acquired by the pupil in the same natural way in which human beings learn and acquire new meanings outside of a textbook. To be more specific, this means that the exact logical sequence of the mature mathematician, which begins with abstract defini- tions and statements of general principles and then proceeds to their applications and specific uses, must give way to a directly psychological method. " Definitions " and general principles must grow out of the pupil's concrete experience. He must begin with details, particulars, concrete elements, and finally arrive at a generalization or the statement of a general principle (seldom, however, at " definitions " in first-ye^r algebra). Con- trast the introductory lesson in algebra which explains that "Algebra is the science of general numbers" with the one which begins, " It saves time to use abbreviations or letters to represent numbers ; for example, you have used C to stand for ioo, M for iooo,etc." The authors have attempted to visualize Foreword v the pupil's mental background \w. writing each page of the exposi- tion of this text and in the selection and statement of problem material. This represents at least a first step toward the design- ing of textbooks on this psychological criterion. The second criterion. The authors' investigations of the math- ematical experience which pupils need to have to prepare them to meet adequately later quantitative situations have resulted in the elimination of a great deal of the content of the traditional book. The time usually given to the operations on polynomials, special products, factoring, complicated fractions, etc., can have no justification in terms of this criterioji of social wo?'tk or utility. The application of this criterion demands a complete recognition of the graphic method of representing number as one of the three methods around which a course should be constructed, and that the formula, the equation, the properties of the more im- portant space forms, and the principle of dependence or func- tionality should form the basic material of the course. Likewise three chapters of the text are devoted to a non-demonstrative study of the triangle. Pupils are shown that mathematics sup- plies notions and devices which people need to master in order to solve many practical problems. The use of scale drawings, angular measurement, the principle of similarity, and the simple trigonometric functions of a right triangle have infinitely more value either (i) from the standpoint of their use in other situa- tions, (2) from their appropriateness and adaptability to the child's interests and abilities, or (3) from the criterion of think- ing value, than does the excessive formalism and manipulation of symbols which they supplant in this course. To summarize, the authors have been concerned to incorporate here the most important mathematical notions which investigation shows that all pupils ought to know. The third criterion. Experimental investigation of the possi- bility of developing powers of generalization shows that such are to be developed only by so complete an organization of courses as will provide a maximum of opportunity for problem-solving, vi Foreword for reflective thinking, for calling into play mental processes of analysis, comparison, and recognition of relations between the parts of problems. Unfortunately, algebra courses have been deprived of most of their " training " value. An emphasis upon formalism, drill, the routine practice in manipulation of mean- ingless symbols, and lack of genuine motive are typical examples of the way in which we have hampered teachers in the develop- ment of problem-solving abilities. The general practice of devoting 80 per cent of the problem-material to these formal drill examples, leaving only 20 per cent for the verbal problem, — which of all types provides most completely opportunity for "thinking," — has been radically modified in the course sub- mitted in this book. The entire course has been organized around a central core of " problem-solving. " Even the purely formal materials themselves have been so organized, wherever possible, as to provide an opportunity for real thinking and not mere habit formation. THE PRIMARY FUNCTION OF MATHEMATICAL INSTRUCTION The writers' thesis in constructing this course of study is this : The central element in human thinking is seeing relationships clearly. In the same way the primary function of a high-school course in mathematics is to give ability to recognize verbally stated relationships between magnitudes, to represent such relationships economically by means of symbols, and to determine such relation- ships. To carry out this aim the course of study, therefore, should be organized in such a way as to develop ability in the in- telligent use of the equation, the formula, methods of graphic repre- sentation, and the properties of the more important space forms in the expression and deter??iination of relationships. Thus we may summarize the chief characteristics of this text, which has been organized on this fundamental aim, as follows : (1) A marked decrease in the emphasis upon formal manipu- lation. The whole course is aimed at providing an opportunity for problem-solving. Foreword vii (2) Three methods of representing number are recognized : the tabular method, the graphic method, and the equational or formula method. Thus graphic representation is an integral part of the course and is not treated as an isolated operation. (3) A vast amount of useless material has been omitted, — for instance, addition, subtraction, multiplication, and division of polynomials ; the more complicated work with fractions, all but the simpler work with radicals, etc. (4) All material shown by investigation to have social utility, and which is omitted from current courses in mathematics, has been included. For example : construction and evaluation of formulas, emphasis on " evaluation " as a most important opera- tion, trigonometric functions of the right triangle, a very com- plete study of variation or dependence, etc. (5) The exposition of the text develops so gradually in accord- ance with the writers' discoveries concerning " learning " that the average pupil can read any part of the discussion and then solve the problems unaided. Thus the text develops in accord- ance with the natural development of the pupil's method rather than that of the highly trained, logical mathematician. This is the outcome of an original analysis of the psychology of mathe- matics. (6) The equation is emphasized throughout as the primary formal operation of the course, — not as an end in itself, but because it is the essential tool for stating and determining quan- titative relationships. THE TEACHER'S TESTS FOR A TEXTBOOK The authors would propose to teachers as tests of the value of material contained in this or in any other textbook the three following principles : (1) Is the type of subject matter presented here or in any other textbook organized in terms of the way children naturally learn ; that is, has the psychological criterion been kept constantly and adequately in mind? (2) Is the pupil who takes this or any other course prompted to do real and viii Foreword genuine thinking ? Does he have ample opportunity for practice in "problem-solving"? Is the subject matter of the course organized primarily around a core of problem-solving situations ? 1 )oes the kind of subject matter presented in this or in any other book sufficiently justify itself from the point of view of its use or importance either in later mathematics courses, in other school subjects, or in situations outside the school ? • The application of these criteria in the construction and selection of school textbooks will go far to bring about the type of reconstruction for which the writers' investigations show there is a real demand. Harold O. Rugg Chicago, Illinois John R. Clark July 25, 1918 CONTENTS CHAPTER PAGE I. A New Way to Represent Numbers . . i II. How to Construct and Evaluate Formulas. 17 III. How to Use the Equation .... 29 IV. How to Represent the Relationship between Quantities w t hich Change Together . . 45 V. How to Find Unknown Distances by Means of Scale Drawings : The First Method . 64 VI. A Second Method of Finding Unknown Dis- tances : The Use of Similar Triangles . 86 VII. How to Find Unknowns by Means of the Ra- tios of the Sides of the Right Triangle . 97 VIII. How to Show the Way in which One Varying Quantity Depends upon Another. . .119 IX. The Use of Positiye and Negative Numbers 130 X. The Complete Solution of the Simple Equa- tion . . . . . . . -152 XI. How to Solve Equations which Contain Two Unknowns . . . . . . .183 XII. How to Solve Equations with Two Unknowns {Continued) . . . . . . .198 XIII. How to Find Products and Factors . .210 XIV. How to Solve Equations of the Second De- gree ........ 232 XV. Further Use of the Right Triangle : How to Solve Quadratic Equations which Contain Two Unknowns 254 FUNDAMENTALS OF HIGH SCHOOL MATHEMATICS CHAPTER I A NEW WAY TO REPRESENT NUMBERS Section 1. It saves time to use abbreviations and let- ters, instead of words, to represent numbers. In order to save time in reading and writing numbers in your studies in arithmetic, you have already found it convenient to use certain abbreviations or letters to represent numbers. For example, instead of " dozen" you have used " doz." to stand for 12 ; C to stand for 100 ; M for 1000 ; cwt. (hundred- weight) for 100 lb., mo. (month) for 30 days, etc. It is necessary that we learn more about this new way of repre- senting numbers by letters because we shall use it in all our later work in mathematics. EXERCISE 1 PRACTICE IN USING ABBREVIATIONS AND LETTERS TO REPRESENT NUMBERS 1. How many eggs are 6 doz. eggs and 2 doz. eggs ? 2. How many days in 3 mo. and 2J mo. ? 3. Change 5 yr. and 2 mo. to mo. 4. If 1 ft. = 12 in., change 4 ft. + 5 in. to in. 5. If R (ream) stands for 500 sheets of paper, how many sheets in 2 R + 3 R ? 6. Change 5 yd. — 2 ft. — 3 in. to in. 7. Using d for 12, how many eggs in 2 d + 3 d eggs ? 2 Fundamentals of High School Mathematics 8. How many sheets of paper in 5R + 3 R — 6 R sheets ? 9. Change 5;;/ + 4w-6w + m to smaller time units, that is, to " days/' if m = 30 days. 10. Change 1 y (yards) + 6/ (feet) to smaller units, that is, to " inches." 11. How many cents in 4 q + 6 d> if q and d stand for the number of cents in a quarter and dime respectively ? 12. If h (hour) equals 60 m (minutes), and m equals 60 s, (seconds), how many s in 2 k +3m? In these examples, you have used abbreviations or single letters to represent numbers or known quantities. We make use of abbreviations or single letters instead of •- because You State the reason here need a great deal of practice in doing this. The next ex- ercise will give more practice in representing numbers by letters in different kinds of examples. EXERCISE 2 FURTHER PRACTICE IN USING LETTERS FOR NUMBERS 1. Change 10 y + 4/ to inches (smaller units) if y and / stand for the number of inches in a yard and in a foot, respectively. 2. Express 2/i + 5 m in seconds. 3. What will 4 R + 3 R + R sheets of paper cost at \$ per sheet? A New Way to Represent Numbers 4. 5. At 4^ each, what will 1 d —2d eggs cost? 10. 11. 12 3f The length of the rec- tangle in Fig. 1 is represented by the expression 3f, and the 2 f width by the expres- sion 2/. What ex- pression will represent the perimeter? How many inches in the perimeter Uf = 12 inches? 6. Change 1 p + 5p — 8p + Sp to ounces, if p stands for the number of ounces in one pound. 7. The expression 14 y + 8 m + 5 d represents the age of a pupil in an algebra class. Express this pupil's age in days or as a certain number of d. 8. Find the cost of 4^ T of coal at 30 cents per cwt., using the relation, T = 20 cwt. 9. If d = 4 q, find how many q in 3 d -f- 5 d. li y = 12 m, 2y + 4 /^ ? and in = 30 d, how many ^/ in A boy earned 27 dollars in a month ; his father earned n dollars. How many dollars would both earn in 3 months, if n stands for 60 dollars? If r= 3 t and /= 6 s, how many s in 4 r — 5/? 13. Write the relation between y and m (*".*. year and month); between / and i (foot and inch); between T and cwt (ton and hundredweight); between h } m y and s {i.e. hour, minute, and second). 4 Fundamentals of High School Mathematics In this you always express " how many" of one unit equals a "certain number" of the other unit. For example, 1/= 12 inches, or, more abbreviated, Are yards, feet, and incites related units ? Are hours, minutes, and seeonds related units ? Are pounds and dollars related units ? Are dollars and cents related units ? Section 2. Word statements about quantities may be much more briefly expressed by using a single letter to represent a quantity. In the last section we saw that it saved time to use abbreviations or letters to represent quantities. Now we shall show that entire word state- ments about quantities may be expressed much more briefly by using a single letter to represent a number. To illustrate, consider next the four different ways of writing the statement of the same example. Illustrative example. (a) The "word" method of stating the example. (b) An abbreviated way to write it. (c) A more abbreviated way to write it. (d) The best way to write it. (a) There is a certain number such that if you add 5 to it the result will be 18. What is the number ? (&) What no. plus 5 equals 18? (c) No. + 5 = 18. (d) n + 5 = 18. It is clear that in all these cases the number is 13, and that it is most easily represented by the single letter n. Thus, the fourth method, n + 5 = 18, illustrates the very A New Way to Represent Numbers 5 great saving that is obtained by the use of single letters for numbers. This method will be used throughout all our later work. This is one of the aims of mathematics : to help you solve problems by better methods than you knew in arithmetic. EXERCISE 3 Express the following word statements in the briefest possible way, using a single letter to represent the quantity you are trying to find. Illustrative example. If a certain unknown number be increased by 7, the result will be 16. This may be most easily written : n + 7 = 16, n = d. 1. There is a certain number such that if you add 12 to it, the result will be 27. What is the number ? 2. If John had 7 more marbles, he would have 18. How many has he ? 3. If the length of a rectangle were 5 inches less, •it would be 21 inches long. What is its length ? 4. A certain number increased by 12 gives as a sum 35. What is the number ? 5. The sum of a certain number and 7 is 18. Find the number. 6. Three times a certain number is 21. Find the number. Explanation : Again, to save time, we agree that 3 times n (8 times p, or 12 times x, etc.) shall be Fundamentals of High School Mathematics written 3 • n or, more briefly, 3 n. Understand, therefore, that whenever you meet expressions like 8/>, 12 x, 17 y, etc., they mean multiplication, even though no " times'' sign ( x ) is printed. In the same way ~ of a certain number is written nor n ; of a certain number, - or - n. 3 3 2 2 2 7. Two thirds of a certain number is 10. Find the number. 8. Three fourths of a certain number is 15. Find the number. 9. The difference between a certain number and 5 is 9. What is the number ? 10. The difference between a certain number and 12 is 13. What is the number? 11. The sum of 16 and a certain number is 29. Find the number. 12. Three fifths of a certain number is 27. What is the number ? 13. The product of 11 and a certain number is 77. Find the number. 14. If your teacher had $50 less, he would have I 15. How much has he ? 15. The quotient of a number and 7 is 3. What is the number ? 16. If the area of a rectangle were increased 12 sq. ft., it would contain 40 sq. ft. What is its area ? A New Way to Represent Numbers 7 17. 13 exceeds a certain number by 4. What is the number ? Explanation : Does this mean that 13 is larger, or smaller, than the certain number ? How do you determine how much larger one number is than another ? 18. Two thirds of the number of pupils in a class is 28. How large is the class ? 19. Tom lacked $7 of having enough to buy a $50 Liberty Bond. How much did he have ? 20. Three times a certain number plus twice the same number is 90. Find the number. 21. The difference between 20 and a certain num- ber is 4. What is the number ? Explanation: It seems most consistent to inter- pret, " the difference between two numbers," as mean- ing, " the first number minus the second number." 1 22. The number of pennies Harry has exceeds 30 by 7. How many has he ? 23. The sum of two numbers is 40. One of them is 27. What is the other ? 24. The difference between two numbers is 21. The larger is 60. What is the smaller ? 25. The product of two numbers is 95. One is 5. What is the other ? 26. The quotient of two numbers is 13. The di- visor is 5. Find the dividend. In these exercises you have been changing, or translat- ing, from the language of ordinary words into algebraic language ; you have been making algebraic statements out 8 Fundamentals of High School Mathematics of word statements. The essential thing in this transla- tion is the representation of numbers by letters. We should note carefully also that we have begun the practice of using a letter to stand for a number which is unknown. ( )i course in these simple examples there is only one un- known number and it is easy to see at a glance each time what it is. HOW TO REPRESENT TWO OR MORE UNKNOWN NUM- BERS, WHEN THEY HAVE A DEFINITELY KNOW RELATION TO EACH OTHER Section 3. In the examples which you have worked in preceding lessons, you have had to represent only one number in each problem. To illustrate, in Example 9, Exercise 3, as in all the other examples solved thus far — "the difference between a certain number and 5 is 9." Only one number has to be represented. But in most of the examples that you will meet in mathematics you will have to represent two or more numbers which have a definitely known relation to each other. For example, consider this problem : Suppose Tom has 5 times as many marbles as John has. How many do they both have ? It is clear that there are two numbers to be represented; namely, the number that Tom has and the number that John has. Furthermore, since there is a definite relation between these two numbers, that is, one is 5 times the other, it is important to see that each can be represented by the use of the same letter. If you let n stand for the number John has, what must represent the number Tom has ? Since the example states * A New Way to Represent Numbers g that Tom has 5 times as many as John, then Tom must have 5 n marbles. In the same way, together they have the sum of the two ; namely, n + 5 n, or 6 n. The best way to state this, however, in algebraic lan- guage, is to use a set form like the following : Let n = the number John has. Then 5 n = the number Tom has, and 5 n -f n, or 6 n = the number both have. The next exercises will show how two or more unknown numbers may be represented by using the same letter, if the numbers have a definite relation to each other. EXERCISE 4 1. Harry has four times as many dollars as James has. If you let n stand for the number of dol- lars James has, what will stand for the number Harry has ? for the number they together have ? 2. The number of inches in a rectangle is 7 times the number in its width. If n stands for the number oi inches in its width, what will repre- sent the number in its length ? in its perim- eter ? 3. An agent sold three times as many books on Wednesday as he sold on Tuesday. Represent the number sold each day. State algebraically that he sold 28 books during both days. 4. There are twice as many boys as girls in a certain algebra class. If there are ;/ girls, how many boys are there? How many pupils? 10 Fundamentals of High School Mathematics State algebraically that there were 36 pupils in the class. Find the number of boys. 5. Oil a certain day Fred sold half as many papers as his older brother. How can you represent the number each sold ? the number both sold ? 6. During a certain vacation period there were three times as many cloudy days as clear days. Express the number of each kind of days, and the total number of days. If the vacation con- sisted of 60 days, how many days of each kind were there ? 7. A rectangle is three times as long as it is wide. If it is x feet wide, how long is it ? What is its perimeter ? 8. If one side of a square is s inches long, what is the perimeter of the square ? State algebraically that the perimeter is 108 inches. 9. The sum of three numbers is 60. The first is three times the third, and the second is twice the third. If n represents the third number, what will represent the first ? the second ? their sum ? State algebraically that the sum is 60, and then find each number. Why do you think it was advisable to represent the third number by n ? 10. John sold five times as many papers as Eugene. If n represents the number Eugene sold, what will represent the number John sold ? What expression will represent the difference in the number sold? Make a statement showing that this expression is 80. A New Way to Represent X timbers n 11. A farmer sold four times as many dollars' worth of wheat as of corn. If he received x dollars for the corn, what will represent the amount he received for both ? 12. A has 11 dollars. B has three times as many as A, and C has as many as both A and B. What will represent the number of dollars all three together have ? 13. A horse, carriage, and harness cost 6500. The carriage cost three times as much as the har- ness, and the horse twice as much as the car- riage. If you let n represent the number of dollars the harness cost, what will represent the cost of the carriage ? of the horse ? of all together? Make an algebraic statement show- ing that all three cost 6500. Can you now find the cost of each ? 14. A man had 400 acres of corn and wheat, there being 7 times as much corn as wheat. Show how the number of acres of each could be rep- resented by some letter. Make an algebraic statement showing that he had 400 acres of both. 15. The rectangle shown in Fig. 2 is three times as long as wide. State al- FlG 2 gebraically that the per- imeter is 64 in. What are its dimensions ? In the problems just studied you have been considering two or more numbers which had a definite RELATION to each other and each of which had to be represented by using the same letter. For example, you had to note that 12 Fundamentals of High School Mathematics one number was always a certain number of times another one, or was a certain part of another one. In each prob- lem you had to decide which of the unknown numbers you would represent by that letter. In general it is best to represent the smaller of the unknown numbers by n or by p or by any letter. The other numbers must then be represented by using the same letter which you selected to represent the first one. EXERCISE 5 Write out the solution of each of the following. Be sure to use the complete form illustrated below, in solving each example. Illustrative example. The larger of two numbers is 7 times the smaller. Find each if their sum is 32. Let s = smaller no. Then 7 s = larger no. and 7 s + s = 32, or 8 s = 32, or 5 = 4, and 7 5 = 28. 1. William and Mary tended a garden, from which they cleared $72. What did each receive if it was agreed that William should get three times as much as Mary? 2. The perimeter of a rectangle is 48 inches. Find the dimensions if the length is 5 times the width. 3. The sum of three numbers is 60. The second is twice the first, and the third equals the sum of the first and second. Find each. 4. Divide $48 between two boys so that one shall get three times as much as the other. A New Way to Represent X umbers 13 5. Twice a certain number exceeds 19 by 5. Find the number. 6. The product of a certain number and 5 is 35. Find the number. 7. A man is twice as old as his son. The sum of their ages is 90 years. Find the age of each. 8. The sum of three numbers is 120. The second is twice the first and the third is three times the first. Find each. 9. The perimeter of a certain square is 14-4 inches. Find the length of each side. 10. The perimeter of a rectangle is 160 inches. It is three times as long as it is wide. Find its dimensions. 11. William is three times as old as his brother. The sum of their ages is 36 years. How old is each ? 12. One number is five times another. Their dif- ference is 10. Find each. 13. The sum of three numbers is 14. The second is twice the first, and the third is twice the sec- ond. Find each number. 14. One number is eight times another. Their dif- ference is 60. Find each. 15. A rectangle (Fig. 3) ~ ^ which is formed by placing two equal squares together has a perimeter of 150 feet. Find the side ~ 7C "~~xT of each square, and Fig. 3 the area of the rectangle. X 14 Fundamentals of High School Mathematics 16. Three men, A, B, and C, own 960 acres of land. B owns three times as many acres as A, and C owns half as many as A and B together. How many acres has each ? 17. John sold half as many thrift stamps as Harry sold ; Tom sold as many as both the other boys together. Find how many each sold, if all sold 144 thrift stamps. 18. Divide #21 among three boys, so that the first boy gets twice as much as the second, and the second boy gets twice as much as the third boy. Section 4. The most important thing in mathematics : the EQUATION. In all the examples in Exercise 5 you have translated word sentences into algebraic sentences. These algebraic sentences are always called equations. They are called equations because they show that one number expression is equal to another number expression. For example, you have stated that n + 5 = 18. This statement merely expresses equality between the number expression n + 5, on the left side of the = sign, and the number expression 18, on the right side. Furthermore, you have been finding the value of the un- known number in each of these equations. From now on, instead of saying "find the value of the unknoivn in an eqtiaiion" we shall say : " solve the equation." For example, if you solve the equation 7s + s = S2, you " find the value " of s ; namely, s = 4. A New Way to Represent Numbers 15 EXERCISE 6 Solve the following Equations : 1. ^> + 3=8. This might be written: What no. + 3=8, or ? + 3 = 8. 2. x — 5 = 10. This might be written : What no. - 5 = 10, or ? - 5 = 10. 3. 2 n = 25. This might be written : 2 times ? = 25. 4. 5 a = 275. This might be written : 5 times ? = 275. 5. l*- = 7. This might be written: \ times ?=7. 6. § c = 12. This might be written : | times ? = 12. It is always helpful to think of an equation as asking a question. Thus, 5 a + 1 = 16 should be thought of as the question : 5 times what number plus one gives 16 ? 7. 2£ + l = 21 14. 3*+ 1=16 8. 5^-3 = 27 15. 7£-2 = 12 9. 4* = 13 16. 12^ = 27 10. 12 = 3/ 17. 16 = 5j/-l 11. 5 + * = 11 18. 6/ + 3/ = 27 12. 6 — « = 2 19. 13 = 5 y 13. 2/ + 3/ = 35 20. 21 = 5.r+l 21. 4* =17 Section 5. Translation from algebraic expressions into word expressions. In the previous work you have trans- lated from word statements into algebraic expressions. It is also very helpful to translate the algebraic expressions back into word expressions. For example, n + 1 = 13 is the 16 Fundamentals of High School Mathematics same as the word statement " the sum of a certain number and 4 is 13." In the same way, the algebraic statement 4 y = 2(> should be translated as follows : 11 the product of a certain number and 4 is 20," or "four times a certain number equals 2b\" The next exercise will give practice in this important process, i.e. translating from algebraic statements into word statements. EXERCISE 7 Translate each of the following algebraic statements into word statements : 1. 2. y + 4 = 20 2/; + l = 31 7. | + 1 = 8 o 12. 13. ^-4 = 12 a-b = l 3. 4. 13 = 2+jy 2tf + 3# = 55 8. 9. 5x = 18 14. d 5. I n + 1 = 12 10. bh = 20 15. \*i-\n = 6. ;/ + 3// = 24 11. h = 60 m SUMMARY OF CHAPTER I After studying this chapter you should have clearly in mind : 1. It saves time to represent numbers by letters. 2. Worded problems may be translated into algebraic statements. 3. Equations are statements that two numbers or two algebraic expressions are equal. 4. Solving equations means finding the value of the unknown number or letter in the equation. 5. Algebraic expressions may be translated into word expressions. CHAPTER II HOW TO CONSTRUCT AND EVALUATE FORMULAS Section 6. Further need for abbreviated language : Short- hand rules of computation. In this chapter we shall study abbreviated or shorthand rules for solving problems. People who have found it necessary to compute over and over again the areas or perimeters of such figures as rec- tangles, triangles, circles, etc., have found it very conven- ient to abbreviate the rules for solving these problems into a kind of shorthand expression which can be more easily writ- ten or spoken than the long rules. For example, suppose you wanted to make a complete statement, either in writing or orally, concern- ing how to find the area of the rectangle which is represented by Fig. 4. You might express it, as you did in arithmetic, as follows : (1) The number of square units in the area of a rectangle is the number of units in its base times the number of units in its height. This long word rule can be greatly shortened by using abbreviations or suggestive letters to represent the number of units in each of its dimensions. Thus, a shorter way of expressing this rule is : (2) Area = base x height. A third and still more abbreviated way of expressing it is : (3) A = b x //, in which A, b, and // mean, respectively, the number of units in the area, base, and height. And finally, remem- bering that b X h is usually written as bh y the entire state- ment becomes : (i) A = b/i. 17 18 Fundamentals of High School Mathematics This last statement tells us everything that the first statement did, and requires much less time to read or to write. Such algebraic expressions are called formulas. Section 7. What is a formula? From the previous illus- tration we see that a formula is a shorthand, abbreviated rule for computing. We must remember, however, that the formula A = bJi is, at the same time, an equation. Since it is an equation that '^frequently used, and which always appears in that particular form, we have come to call it a FORMULA. I. COMPUTATION OF AREAS AND PERIMETERS BY FORMULA EXERCISE 8 COMPUTATION OF THE AREA OF RECTANGLES BY THE FORMULA l. Illustrative example. Find the area of a rectangle (Fig. 5) in which b= 10 and h = 7.5, using the formula A = bh. Solution : (1) A = bh. (2) A= 10 x 7.5. (3) A = 75. 2. Find A when b = 8.25 and h = 4. 3. Find A when h = 4.5 and b = 12. 4. Find b when A = 50 and h = 5. 5. What is A if b = 6.5 and h = 5.4 ? 6. What is h if A = 40 and b = 6| ? 7. Find A if h = 2.5 and b = 6.4. 8. What is b if A = 450 and // = 22.5 ? How to Construct and Evaluate Formulas 19 9. If A = 200 and b = 7.5, what does h equal ? 10. If A = 625 and h = 50, what does b equal ? 11. What is ^4 if b = 40 and h is twice as large as b ? 12. Find ^ if h = 16.2 and * = \ of A. 13. b == 12 and ^ = f £. What is ^ ? 14. Find ^ if h = 20 and A + b = 32. Section 8. Perimeters of rectangles. In Section 7 we saw that it was convenient to use a formula for the area of a rectangle. In the same way it is helpful to have a formula for the perimeter of any rectangle. Since' the perimeter of any rectangle is the sum of the bases and altitudes, the shortest way to ... J Fig. 6 express this is : (1) Perimeter = 2 x base plus 2 x height, or by the formula (2) P = 2 b + 2 h. EXERCISE 9 COMPUTATION OF THE PERIMETERS OF RECTANGLES BY THE FORMULA 1. Illustrative example. Find the perimeter if the base is 13 and the height is 9 ; or, more briefly, find P if b = 13 and h = 9. Solution: (1) P=2b + 2h. (2) P=2-13 +29. (3) P = 26 + 18 = 44. 2. Find P if h = 10.5 and * = 9. 3. Find P if >fc = 18.4 and b = 12.8. 4. What is kiiP = 40 and b = 10 ? 2o Fundamentals of High School Mathematics What is b if P = 60 and h = 14? Find A if />= 18.4 and * = 4.6. If P = 110 and A = 22.5, what is b? What is P if h = 18 and * = 2 A? /> = 1 00. Find * and h if * = //. What is h\lP = 120 and * = £ P ? 5. 6. 7. 8. 9. 10. Section 9. The formula for the area of any triangle. What is the area of this triangle if its base is 12 ft. and its height is 8 ft. ? How do you find the area of any triangle if you know its base and altitude ? Show that the most economi- cal way to state this rule, or re- lation, between the area, the base, and the height, is by the formula A = — . FlG# 7 The examples in the following exercise will give you prac- tice in using this important formula. EXERCISE 10 COMPUTATION OF THE AREA OF TRIANGLES BY THE FORMULA l. Illustrative example. Find the value of A if b = 22 and h = 12. Solution: (1) A = — . v 2 (2)A = 22 x 12 264 = 132. Write your work in a neat, systematic form. 2. Find the value of A if b = 18 and h = 6|. 3. What is the value of A if b = 12.5 and // = 20? How to Construct and Evaluate Formulas 21 4. If h = 16.8 and b = 28, what does A equal ? 5. What is the value of b if A = 300 and h = 50 ? 6. Find h if A = 240 and * = 20. 7. Determine £ if ^4 = 100 and h = 15. 8. What is AHA = 6.25 and * = 10.5 ? 9. Find the value of A if b = 22 and /* = T 6 T £ 10. What is b if ^ = 120 and h = \ A ? 11. /z = 20 and b — \k. What does ^ equal? 12. Can you find b and // if ^4 = 256 and b = 2/i? 13. Two triangles have equal bases, 10 in. each, but the height or altitude of one is twice that of the other. Are their areas equal? Show this by an illustration. 14. What change occurs to A if b is fixed in value, but if h gets larger ? What is the relation be- tween A and h if b is fixed ? Section 10. The formula for the circumference of any circle. You will recall from arithmetic the following state- ment for the circumference of a circle : " The circumference of a circle is obtained by multiplying twice the radius by 3.1416." With our new method of using letters instead of words or numbers, this is much more briefly expressed by the formula FlG - 8 C = 2 irR. 7r is a symbol used to represent the number 3.1416. Use this value for it in the problems which follow. 22 Fundamentals of High ScJiool Mathematics I EXERCISE 11 COMPUTATION OF THE CIRCUMFERENCE OF CIRCLES BY THE FORMULA 1. What is the value of C if R = 12 ? 2. What does C equal if R = 5| ? 3. Find the value, of R if C= 31.416. 4. Determine R when C= 100. 5. Find C if the diameter of the circle is 16.4. 6. What is the value of C if R = ^ ? 7. A Boy Scout wishes to make a circular hoop from a piece of wire 14 ft. long. Determine the radius of the largest possible hoop he can make. 8. The radius of one circle is 5, and the radius of another is twice as great. Find the circumfer- ence of each, and note whether one circumfer- ence is twice the other. 9. Think of a circle of some particular radius. Then imagine that the radius begins to increase. What happens to the circumference ? Is there any particular connection or relation between C and R ? II. "EVALUATION": HOW TO FIND THE NUMERICAL VALUE OF AN ALGEBRAIC EXPRESSION Section 11. In the examples which you have just solved wc have used the long expression " What is the value of" or " Find the value of " in referring to the particular letter which was to be found. Instead of these long expressions we shall now use the single word evaluate. It means exactly the same thing as the longer expression. Thus How to Construct and Evaluate Formulas 23 to evaluate an algebraic expression means to find its nu- merical value, exactly as in the previous examples. This is done by " putting in" or by substituting numerical values for the letters. A few examples will make this clear. EXERCISE 12 EVALUATION OF COMMONLY USED FORMULAS 1. Evaluate A = ^ if b = 10 and h = 14.6. 2. Evaluate, or find the value of, P in the expres- sion P = 2 6 + 2 h if £ = 26 and h = 12.4. 3. Evaluate C = 2 irR if R = 14. 4. Evaluate F = /ze>// if /= 10, w = 6J, and // = 5. 5. Find the value of i in the formula i=prt if / = $640, r = yo"o> an d * = 4. 6. Evaluate A = ttT? 2 if R = 6. 7. Evaluate ^ = - if E = 110 and i? = 10.5. R 8. What is h in the algebraic expression P = 2d + 2kifP=8Q and £ = 12.8? III. THE USE OF EXPONENTS TO. INDICATE MULTIPLICATION Section 12. Need of short ways to indicate multiplica- tion. A very large part of our work in mathematics is that of finding numerical values. In many of our problems, therefore, we shall need short ways of indicating multipli- cation. For example, in arithmetic, the multiplication of 5 x 5 is sometimes written as 5 2 ; or the multiplication of 24 Fundamentals of High School Mathematics 6 x 6 x 6 as 6 8 . In algebra, to save time, this notation, or method of indicating multiplication, is always used. Thus, instead of writing" b X b or u x u X u we will write b 2 or /z 3 . This little number that is placed to the right of and above another number tells how many times that number is to be used as a factor. These numbers are called exponents. Numbers with exponents are read as follows : 3 a 2 means 3 times a times a, and is read "3a square.' 11 This does NOT mean 3 a times 3 a. The exponent affects only the a. 5 #* means 5 times b times b times b, and is read " 5 b cube." This does NOT mean 5 b times 5 b times 5 b. The exponent affects only the b. Here, as well as throughout all later mathematical work, you will need to be able to evaluate algebraic expressions which involve exponents. For example, the area of the rec- tangle shown here is the expres- sion 3 IV 2 , which is obtained by multiplying 3 W by W. Now the numerical value of this area depends upon the value of W ; that is, if W is 4, then the area is 3-4-4, or 48 ; but if W is 2, then the area is 3-2-2, or 12. In the same way the volume of the rectangular box in Fig. 10 is represented by the ex- pression 2x s , or 1x • x • x. Again, you see that the numerical value of the volume depends upon the value of x. Thus, if x is 5, the volume is obtained by evaluating the expression 2x s , which gives 2 • 5 • 5 • 5, or 250. The next exercise gives practice in evaluating algebraic expressions containing exponents. -3W- T Fig. 9 Fig. 10 How to Construct and Evaluate Formulas 25 EXERCISE 13 PRACTICE IN EVALUATION 1. Illustrative example. Evaluate 2 a& 2 + 3 a 2 b + ac, if a — 4, b = 3, and c = 1. Solution : 2- 4- 3- 3 + 3-4. 4- 3 + 4-1 = 72 + 144 + 4 = 220. Note that the numbers are substituted for, or put in place of, the letters. a + b — c 9. aHc* 10. M+ 1 a o c 11. a s — b s — c z 2 a 2 12. a h + b a + c a Using the values of a, b, and c given in Example 1, evaluate each of the following expressions : 2. a 2 +d 2 + c 2 a-b + c 3. 3 abc 4. a 2 b + ab 2 5. ac 2 + cb 2 + ba 2 6. a 3 + b z + c 3 7. ? + b - + C - b a b 13. The formula d= 16 / 2 tells how far an object will fall in any number of seconds. Find how far a body will fall in 1 sec. of time, that is, when t = 1, Do you believe it ? How could you test it ? 14. Using the formula in Example 13, find how far an object will fall in 2 sec. of time, that is, when t = 2. How could you test the truth of this? 15. The horsepower of an automobile is given by ZM 72 the following formula: -^~~y in which D rep- 2.5 resents the diameter of the piston, and N the 26 Fundamentals of High School Mathematics number of cylinders. What is the horsepower of a Ford, which has 4 cylinders, and in which D = 31 in. ? IV. THE CONSTRUCTION OF FORMULAS Section 13. It is very important to be able to make a formula for any computation that must be performed over and over again. For example, we often have to find the area of a square. Instead of saying or writing each time " the area of a square is equal to the square of the number of units in one of its sides," it saves time to use the for- mula A = s 2 , in which A = area and s = one of the sides. This formula tells all that the word ride says arid requires much less effort. To give practice in this kind of work, construct a formula for each of the examples in the following exercise. EXERCISE 14 1. (a) Find the volume of a rectangular box whose dimensions are 12, 8, and 6 inches. (b) Make a formula for the volume of any box. 2. (a) What is the area of a circle whose radius is 9 in. ? (b) Write the formula for the area of any circle. 3. {a) How many square inches in the entire sur- face of a cube whose edge is 8 inches ? (b) Give a formula for the area of the entire sur- face of any cube. 4! (a) Find the interest on %400 for 2 years at 6%. (b) Make a formula for the interest on any prin- cipal for any rate and for any time. How to Construct and Evaluate Formulas 27 5. (a) How many cubic inches in a block 2 1 by 3' by 4'? (b) Make a formula for the number of cubic inches in any rectangular solid whose dimen- sions are expressed in feet. 6. Make a formula for, or an equation which tells, the cost of any number of pounds of beans at 12 cents per pound. 7. What equation or formula will represent the area of any rectangle whose base is 5 inches, but whose height is unknown ? Evaluate your for- mula for h = 3.4. 8. An automobilist travels 20 miles per hour. What formula or equation will represent the dis- tance he travels in / hours ? Evaluate this formula : t = 5 hr. 20 min. REVIEW EXERCISE 15 1. What does an equation express ? Is 7 + 4 = 6 + 6 an equation ? 2. Does 2 n + 1 = 21, if n = 9, make an equation? if n = 10 ? 3. The formula for the perimeter of a rectangle, p = 2 b +2b y contains three unknown numbers. How many of them must be known in order to use this formula to solve an example ? 4. Read each of the following equations as ques- tions, and find the value of the unknown number: (a) ±y + 3 = 21 (d) 3 y - 5 = 16 (6) 20 = 6 + 2x ( e ) x + x= 36 (c) 5 c + 2 = 42 (/) &+ b + 1 = 23 28 Fundamentals of High School Mathematics 5. Three times a certain number, plus 2, equals 38. Find the number. 6. Donald saved twice as much money as his older brother. Express in algebraic language that both together saved $ 96. How much did each save? 7. Evaluate the formula V=P(V=the volume of a cube), if /= 4.]. 8. The first of three numbers is twice the second, and the third is twice the first. Find each number if their sum is 105. 9. Construct a formula for the cost of any number of eggs at 30 cents per dozen. 10. What is the difference in meaning between 10 n and ;/ + 10 ? Does 4 w mean the same as 4 + w ? SUMMARY The most important principles and methods which we have learned in this chapter are the following : 1. A formula is merely a shorthand rule of computation. 2. Formulas are " evaluated" or "solved" by sub- stituting numbers for the letters in the formula. 3. Exponents are used as short methods of indicat- ing multiplication. An exponent of a number tells how many times that number is taken as a factor. 4. We should construct a formula for any kind of problem which we have to solve frequently. CHAPTER III HOW TO USE THE EQUATION Section 14. The importance of the equation. Nothing else in mathematics is as important as the equation, and. the power to use it well. It is a tool which people use in stat- ing and solving problems in which an unknown quantity must be found. In the last chapter we saw that the formula^ or equation, was used to find unknown quantities, sometimes the area, sometimes the perimeter, etc. The fact that the equation is used as a means of solving such a large number of problems is the reason we shall study it very thoroughly in this chapter. Section 15. The equation expresses balance of numerical values. The equation is used in mathematics for the same purpose that the weighing " scale " is used by clerks ; that is, to help in finding some value which is unknown. The scale represents balance of weights : similarly the equation represents balance of numerical values. To un- derstand clearly the principles which are applied in dealing 3<d Fundamentals of High School Mathematics with equations, we should consider the scale, as repre- sented in Fig. 11. In this case a bag of flour of un- known weight, together with a 5-pound weight, balances weights which total 25 pounds on the other side of the scale. Now, if ;/ represents the number of pounds of flour, it is clear that the equation 7i + 5 = 25 represents a balance of numerical values. Obviously, n is 20, for the clerk wtmld take 5 pounds of weight from each side, and still keep a balance of weights. This principle, namely, that the same weight may be taken from each side without destroying the balance of weights, can be applied to the equation n + 5 = 25. That is, we may subtract 5 from each side of- the equation, giving another equation, ^ = 20. This suggests an important principle that may be used in solving equations ; namely, — The same number may be subtracted from each side of the equation without destroying the equality, or balance, of values. If you take something from one side of the scale, or of the equation, what must you do to the other side ? Why ? The fact that the equation expresses the idea of balance makes it easy to reason about it, and find out all the things that can be done without changing the balance or equality. The next exercise suggests this kind of study of the equa- tion. How to Use the Equation 31 EXERCISE 16 By thinking of the equation as a balance, you should be able to complete the following statements. Fill in the blanks with the proper words. 1. Any number may be subtracted from one side of an equation if 1 is ! from the other side. 2. Any number may be added to one side of an equation if ! is I to the other. 3. One side of the equation may be multiplied by any number, if the other side is I by the 4. One side of an equation may be divided by any number, if the other side is I by the These are very important principles, and are used in solving any equation. They are generally called axioms. They must be understood and mastered. The examples, of the next exercise have been planned to help you learn how to apply them. EXERCISE 17 In each of the following examples you can use one of the four principles stated above to explain what has been done, or to state the reason for doing it. Thus, if 2,r=8, then x = i } because of the principle: " One side of an equation may be divided by a number if the other side is divided by the same number." For each example, you are to state the principle which permits or justifies the conclusion. 32 Fundamentals of High School Mathematics 1. If 4/; = :!:!, then what is done to each side to give b = 5 J ? 2. If ]j' = 7, then to get y = 14, what do you do to each side ? 3. If a - + 4 = 13, then to get x = 9, what do you do to each side ? 4. If 5c = 32. 5, then what is done to each side to give c = 6.5 ? 5. If 6a = 12, then what is done to each side to give 3 a = (> ? 6. If y — 4 = 7, then what is done to each side to give y = 11 ? 7. If x = 2 and j/ = 3, then why does x +y = 5 ? 8. If b = 3 and £ = 10, then why does fo = 30 ? 9. If x = 12 and jj/ = 4, then why does - = 3 ? y 10. If a — 1 = 9, then why does # = 10 ? . ll. If b = 2/i, then what is done to each side to give 3 6 = 6 // ? 12. If -ix + 3 = 23, then what is done to each side to give 4 x = 20 ? 13. If oy — 3 = 27, then what is done to each side to give by = 30 ? 14. If x + 7 = 19, then to make ;r = 12, what is done to each side ? 15. If 2c — 4 = 8, then to make 2^ = 12, what is done to each side ? 16. If 3 b + 1 = 22, then what is done to each side to give b = 7 ? How to Use the Equation 33 17. If 5^ + 2 = 47, why does 5 b = 45 ? Then why does b — 9 ? 18. If 6;r+2 = ;r + 22, then why does 5* + 2 = 22 ? and why does 5;r=20? and why does ;r = 4? 19. If you know that ±w + 3 = w + 27, then why does 4 w = w + 24 ? and why does 3 w = 24 ? and why does w = 8 ? These examples are given to emphasize the fact that there are certain changes that can be made on both sides of an equation, without destroying the balance or equality. It should be clear that there must be some axiom or prin- ciple to justify every change that is made. EXERCISE 18 Find the value of the unknown number in each of the following equations, telling exactly zvhat you do to eacli side of the equation. 1. x + 5 = 13 2. 2, a = 11 3. 26 = 4j' 4. 2£ + l = 19 5 « y -5 = 12 6. J-* = 4.5 7 - a r 3 = 5 8. 2*- 3 = 17 9. 15 = x + 7 10. 2.r + 3;r = 35 11. 5y + -iy+y = 30 12. 15 = 3 b 13. 5 c - 2 = 38 14. 27 = 6.r -3 15. 4 b + 7 = 47 16. 4j-j/ = 21 17. 5x + 1=23 18. 21* = 15 19. f« = 18 20. y 34 Fundamentals of High School Mathematics 21. 2£ + 3*=42 24. 18+* = 13 + 10 22. *r-4 = 13 25. 7 + 2* = 23 + 10 23. 2/;- 1=18 26. £*+£* = 18 HOW TO CHECK THE ACCURACY OF THE SOLUTION OF AN EQUATION Section 16. When is an equation solved ? We have already noted that an cquatioii is solved when the numerical value of the unknown number is found. Thus, the equa- tion 4 a + 3 = 29 is solved when the numerical value of a is found. This leads to another very important question ; that is : How can you be certain your solution is correct ? In other words, how can you test or check the accuracy of your work ? For example, suppose that in solving the equation 4 a + 3 = 29 one member of your class obtains 8 for the value of a. Is his result correct ? There is only one way to be sure. That is to substitute or " put in " 8 in place of a in the equation, to see whether the numerical value of the left side equals the numerical value of the right side. In other words, does 4 . 8 + 3 = 29 ? Clearly, not. Therefore, the solution is incorrect ; it does not check. Then what is the correct value of a ? Some of you doubtless think it is 6|. Let us test or check by substituting 6J for a, to see if the numerical value of one side of the equation will equal the numerical value of the other side. Does 4.6,] +3 = 29? Yes. Then the equation is solved, or, to use the more gen- eral term, the equation is satisfied when a = 6|. How to Use the Equation 35 Summing up, then, an equation is solved when a value of the unknown is found which satisfies the equation ; that is, one which makes the numerical value of one side equal to the numerical value of the other side. The solution of the equation is checked by substituting for the unknown number the value which we think it has. If, as the result of the substitution, we get a balance of values, then we know that the equation has been solved correctly. We have already had practice in substituting numerical values for letters. In Chapter II, we called this evalua- tion. Thus you see that each time you check the solution of an equation, you are evaluating the original equation, EXERCISE 19 PRACTICE IN CHECKING THE SOLUTION OF EQUATIONS 1. The pupils in a class tried to solve the equation 6^-3 = 39. A few decided that a = 7, while the others in- sisted that a = 6. Which group was right ? Show how they could have checked or tested their re- sult. Why, do you think, some pupils got 6 for the value of a ? 2. Does x = 5 in the equation 12 x — 7 = 10 x + 3 ? In other words, does x = 5 satisfy this equation ? 3. Would you give full credit on an examination to a pupil who said that y = 4J would satisfy the equation 8y — 4 = 6 y + 3 ? Justify your answer. 4. Show whether the equation b 2 + 5 b = 24 is satis- fied or solved if b = 3 ; if b = 2. 5. Do you agree that the value of x is 6 in the equa- tion 10.*: — 4 = 58 ? Justify your answer. 36 Fundamentals of High School Mathematics 6. Is the equation - ^ + 6 = j b + 8 satisfied when /> = 8 ? 7. Does -r = 24 satisfy the equation \x+\x + \x = t 2&} 8. State in words how the solution of an equation is tested or checked. 9. What is the value of learning to check very care- fully every kind of work you do ? EXERCISE 20 Solve each of these equations. Write out your ivork for each one in the complete form illustrated in the first example. Check each one so that you can be absolutely certain that your work is correct. 1. Illustrative example. 6 b - 4 = 24. (1) By adding 4 to each side, we get 6 b = 28. (2) By dividing each side by 6, we get b = 4|. (3) Checking, 6 • 4f - 4 = 24. 28 - 4 = 24. 2. 5 * - 2 = 38 10. Sx + 2x+6x = 66 3. 6 £ + 3 = 45 11. 5 c + 3 = 78 4. T x = x + 30 12. 1 v — 8 1 5. 6. 22 = 5* + 2 13. 5 " 4 7. 2.]j/ + l =26 14. 10/; + 3 = 7£+15 8. b + 5 b = 20 + * 15. 12*- 2 = 5*+ 26 9. iy = lZ+y 16. 13 j/ = 2 j/ + 3 j/ +4^/ How to Use the Equation 37 HOW TO GET RID OF FRACTIONS IN AN EQUATION Section 17. The use of the most convenient multiplier. In many equations that you have solved already it has been necessary to multiply each side of the equation by some number. For example, in \x — 10, it is necessary to mul- tiply each side by 2, which gives x = 20. Or, if you wanted to solve the equation 1^ = 3, it is necessary to multiply each side by 5, giving x = 15. But, suppose you had an equation would you get rid of both fractions by multiplying each side by 2 ? Would you get rid of both fractions by multiplying each side by 5 ? Here, as in all equations of this kind, you have to find some number which is a multiple of the different denominators. For this reason, in this example, 10 is the most convenient number by which to multiply each term in the equation. Illustrative example. Multiplying each side of (1) by 10, we get 10. i* -f 10.^x = 10- 14, or 5 x -f 2 x — 140, or 7 x — 140, or x = 20. Checking, by substituting the value of x in the original equation, gives 1.20 + ^20=14, 10 + 4 = 14. The study of this example shows that we can get rid of fractions in an equation if we multiply each side by the lowest common multiple of the denominators. We shall call this the most convenient multiplier. 38 Fundamentals of High School Mathematics EXERCISE 21 PRACTICE IN SOLVING FRACTIONAL EQUATIONS Solve and check each example. 1. Illustrative example. * } n + \ n = 22. What is n ? (1) Multiplying each side by the most convenient multiplier, 12, gives 12 • f n + 12 • J n = 12 • 22, or 8 n + 3 n = 264. (2) By adding 8 n and 3 n, we get 11 n = 264. (3) By dividing each side by 11, we get 72 = 24. (4) Checking, by substituting the value of n (24), in the original equation, 22. 16 + 6 = 22. 2. What is the value of x in the equation \x+\x = m 4. A man spent \ of his income for rent and \ for groceries. Using n to represent his income, make an equation which will state that he spent $ 660 for rent and groceries. Solve the equation. The dimensions of a rectangle are indicated on Fig. 12. What equation will state that the perimeter is 36 in. ? Solve the equation for /. Fig. 12 How to Use the Equation 39 5. 4 b + b = 1 j + 13. What is £ ? 6. If three fourths of a single number be dimin- ished by one half of the number, the remainder is 10. Find the number. 7. 2 X + % X -1 = 3. 8. Three boys together had 65 cents. Tom had half as much as Harry, and Bill had two thirds as much as Harry. Translate this into an equation, and solve. 9. ? + 2 + 5 =4 + ? What does n equal ? 2 3 2 10. One half of a certain number increased by four fifths of the same number gives 52 as a result. Find the number. 11. Harry made two thirds as much money last year selling the Saturday Evening Post as John made ; Edward made three fourths as much as John. How much did each boy earn if all to- gether earned $ 145 ? 12. ^- + |^-6 = 24. 13. The sum of the third, fourth, and sixth parts of a number is 18. Find the number. Section 18. How word problems are solved by equations. It is important to note the principal steps involved in solving word problems. Let us take, as an illustration, Example No. 8 in Exercise 21. Three boys together had 65 cents. Tom had half as much as Harry, and Bill had two thirds as much as Harry. How much had each ? 4-o Fundamentals of High School Mathematics 1. The first important step in solving a word problem is to get in mind very clearly what is known and to recognize what is to be found out. In all problems some things are known and some things are to be determined. Thus, in this problem, we know how much money all the boys have together ; and we also know that Tom has half as much as Harry ; further- more, we know that Bill has two thirds as much as Harry. That is, we see that the statement of the amount that Tom and Bill each has depends upon the statement of the amount that Harry has. 2. But we do not know how much Harry has. Then, as in all word problems, we represent by some letter, such as n, the number of dollars Harry has. In other words, the second step is to get clearly in mind what quantities are un- known, and to represent one of them by some letter. 3. Next, all the parts or conditions of the prob- lem must be expressed by using the same letter. Thus, if Harry has n dollars, the number that Tom and Bill each has must be represented by using the same letter n, and not some other letter. That is, the word statement must be translated into an algebraic statement. It is always necessary, and usually difficult, to see that there must be a balance, an equality, between the parts of the problem. Thus, we must see that Harry's money, n t plus Tom's money, \ n y plus Bill's money, | n, How to Use the Equation 41 must balance, or equal, 65 cents. This gives the complete algebraic statement : n + \ n^\- § n = 65. 4. The equation which we have obtained must be solved. A value of the unknown must be found which will satisfy the equation. In this case n proves to be 30. 5. Finally, the accuracy of the result must be tested by substituting the obtained value of n in the original word statement of the problem, to see if the statement holds true. EXERCISE 22 Translate into algebraic language, and solve each of the following word statements. Check each one. 1. Six more than twice a certain number is equal to 12. Find the number. 2. Four times a certain number is equal to 35 di- minished by the number. What is the number? 3. I am thinking of some number. If I treble it, and add 11, my result will be 32. What number have I in mind ? 4. If fourteen times a certain number is dimin- ished by 2, the result will be 10. Find the number. 5. What is the value of y in the equation 4j/ + i?/ + 2 = 28? 6. If seven times a certain number is decreased by 8, the result is the same as if twice the number were increased by 32. Find the number, 4.2 Fundamentals of High School Mathematics 7. An algebra cost 12 cents more than a reader. Find the cost of each if both cost $1.64. 8. The sum of the ages of a father and his son is 57 years. What is the age of each if the father is 29 years older than the son ? 9. The length of a school desk top exceeds its width by 10 inches; and the perimeter of the top is 84 inches. What are its dimensions? 10. Divide $93 between A, B, and C, so that A gets twice as much as C, and B gets $10 more than C. 11. A farmer sold a certain number of hogs at $20 each, and twice as many sheep at $14 each. How many of each did he sell if he received $576 for all? 12. Should a teacher give James full credit for the solution of the equation 41^-7 = 3^ + 5 if he obtained x=8^? Justify your answer. 13. Make a drawing of a rectangle whose perimeter is represented by the expression 6 y + 20, writ- ing the dimensions on the drawing. 14. The length of a rectangle is 5 inches more than twice its width ; its perimeter is 46 inches. What are its dimensions ? 15. A school garden was 3| times as long as wide. To walk around it required 31 steps (27 in. each). Tell how to find its width, but do not actually find it. How to Use the Equation 43 REVIEW EXERCISE 23 1. The formula /i 2 = a 2 + b 2 is used in finding a side of a right triangle. Evaluate it if the base is 13 and the altitude is 5. 2. Make a formula for the number of revolutions made by the front wheel of a Ford car in going a mile, if the radius of the wheel is 14 inches. 3. In what sense does the equation 7<5 — 5 = ^ + 25 ask a question ? 4. Give one illustration of the advantage of using letters for quantities. 5. What are the four fundamental principles or axioms which are used in solving equations? 6. Does x — \.\ satisfy the equation x 1 — 3 x = 6 ? 7. Using m, s, and d for minuend, subtrahend, and difference, respectively, what equation or equa- tions can you make from them ? 8. An autoist travels at an average rate of 24 mi. per hour. What distance will he cover in 2 hr. ? in 5 hr. ? in 10 hr. ? Make an equation or for- mula for the distance he will travel in t hr. 9. Write a formula for the cost of any number of pounds of bacon at 30 cents per pound. 10. Draw rectangles with bases of 2 inches each. What formula will represent the area of any such rectangle if // represents the height ? 11. How do you get rid of fractions in an equation ? What is the most convenient multiplier in any particular equation ? 44 Fundamentals of High School Mathematics 12. When is an equation solved f 13. If J/, ;;/, and/ stand for the multiplicand, mul- tiplier, and product, respectively, what formulas can you make from them ? 14. If ;/ is a whole number, what is the whole num- ber next larger than n ? the whole number next smaller than ;/ ? 15. In getting rid of fractions in the equation 2 X ' 3 X == » show that 6 is a more convenient multiplier than 12, 24, 18, 30, etc. SUMMARY From your study of this chapter, the following principles and methods should be kept clearly in mind: 1. Equations express balance of value. 2. If any change is made on one side of the equa- tion, the same change must be made on the other side. 3. An equation is solved when a value of the un- known is found which satisfies the equation. 4. The accuracy of your solution is checked by evaluating the equation for the value of the unknown. 5. You can get rid of fractions in an equation by multiplying each side by the lowest common multiple of the denominators ; that is, by the most convenient multiplier. CHAPTER IV HOW TO REPRESENT THE RELATIONSHIP BETWEEN QUANTITIES WHICH CHANGE TOGETHER Section 19. The chief aim of mathematics. As we go about our daily work, we commonly deal with quantities which change together. For example, the cost of a rail- road ticket changes as the number of miles you travel changes ; that is, the cost and the distance change together. Or, the distance traveled by an autoist, if he goes at the rate of, say, 20 miles per hour, changes as the number of hours which he travels changes ; that is, the distance and the time change together. As a third illustration, suppose you wanted to make a trip of 100 miles. We know that the time required will change with or be determined by the way in which the rate changes ; that is, the time and the rate change together. The fact that we are always dealing with situations of this kind makes it necessary for us to know how to represent and determine these quantities which change to- gether, or which are related in some definite way. Mathematics shows us how to describe or express them. In fact, it is the chief aim of mathematics to help you to see how quantities are related to each other and to help you to determine their values. Section 20. The three methods of representing relation- ship. People have used three different methods for doing this: I. THE TABULAR METHOD II. THE GRAPHIC METHOD III. THE EQUATIONAL OR FORMULA METHOD This chapter will show how quantities, which are so re- lated W\2X they change together, can be represented by these methods. (// should be pointed out that it is not always possible to use the equational or formula method.) 45 46 Fundamentals of High School Mathematics To illustrate, suppose you wanted to tell some one about the temperature in Chicago on a certain July day. There are two ways to do this ; first, you might make a table, like the following, which would tell the temperature at each hour during the day. Table 1 an illustration to show the way to tabulate temperature at different hours of the day A.M. P.M. Hour 6 7 8 9 IO n 12 1 2 3 4 S 6 Temperature 70 72 72 7& 80 83 85 90 96 94 86 76 77 This method, which we shall call the tabular method, shows the way the temperature changes at different hours of the day. To understand the table, however, requires much more effort on the part of the reader than is re- quired to understand the second method, which is shown below. This pictorial or graphic method shows all that the tabular mctJiod shows, and has the advantage of being more easily interpreted. _. 1 r ..._ — — _ — 100° " < -^~-~*~ Sw y ' J**^ S - ... S :::::::::::::^-: :: :::::: : 5 — !::: Ill -r n o —"'^ ^ 70 — -^*- 60 -~ 10 12 TIME Fig. 13. Graphic representation of temperatures at various hours of the day. Note that the horizontal line, or scale, shows the hours, or the time ; each large space represents one hour. The How to Represent Relationship between Quantities 47 vertical line, or scale, shows the temperature ; on this scale each large space represents 10°, or each small space repre- sents 2°. Suppose we wanted to read from the graph what the temperature was at 11 o'clock. We find it by- looking along the time line, or time axis, until we come to the point marked 11 o'clock. We then look up, or down, to the line of the graph. In this case we have to go up to a point 11 \ small spaces above the 11 o'clock point. By looking back, to the left, to the vertical or temperature scale, we see that any point on this horizontal line stands for 83°. Hence the graph shows that at 11 o'clock the temperature was 83°. The following questions will help you compare the graphic and tabular methods of representing the relation between two numbers. EXERCISE 24 In order to answer each of these questions, refer to the data of Table 1 and Fig. 13. 1. Find, both from the table and from the graph, the highest temperature. 2. What was the lowest temperature ? Which shows this the more easily, the table or the graph ? 3. Between what hours did the temperature change the most rapidly ? 4. About what do you think the temperature was at 9.30 a.m. ? 5. Between what hours did the temperature change the least ? 6. What might explain the rapid fall in temperature between 4 p.m. and 5 p.m. ? 48 Fundamentals of High School Mathematics After answering the questions, are you not convinced that the graphic method gives the information which the reader may desire much more quickly and easily than the tabular method ? The fact that this is true has brought about a very wide use of graphic methods in all kinds of business and industry. Nearly every newspaper and magazine contains "graphs" of some kind. Your teacher will be glad to have you bring to class any graphs you may find in the newspapers or magazines. EXERCISE 25 1. An east-bound train, running at 40 miles per hour, left Chicago at 8 a.m. Show from the graph, Fig. 11, how far the train was from Chicago at 10 a.m. ; at 11 a.m. ; at 11.30 a.m. ; at 2 p.m. At what time was the train 100 miles from Chicago ? 200 miles ? 250 200 W U lrA £ 150 *r* 2 H V) £ioo : Q 50 m t ?*~ J. s . r\ * <r 8 10 11 12 TIME Fig. 14. The line shows relationship between time spent and distance traveled. How to Represent Relationship between Quantities 49 2. In Fig. 14 how many miles does each small space represent? How many hours does each large space equal ? 3. In a newspaper the following graph, Fig. 15, was printed. It gives the prices of wheat, per bushel, from August 5 to August 10. What was the price on August 5 ? On August 7 ? On August 9 ? $1.40 Wj.30 \, u 120 110 i 9 10 5 6 7 8 AUGUST Fig. 15. Graphic representation of prices of wheat on various days of August, 1916. When was the price the highest ? the lowest ? Between what dates did the price change most ? change least ? Each large space on the vertical scale repre- sents how many cents ? What is measured along the horizontal scale ? What is the unit used on this scale ? SUMMARY OF IMPORTANT ASPECTS OF GRAPHIC REPRESENTATION Section 21. In the study of the previous examples, the following important aspects of graphic representation should be noted : 50 Fundamentals of High School Mathematics 1. Graphs always show the relation between two changing quantities; for example, they showed the relation between the number of miles traveled and the time required. 2. Two rectangular axes are drawn. One of the changing quantities is measured on the hori- zontal axis ; the other changing quantity is measured on the vertical axis. 3. These axes, or reference lines, are scales, marked off in a series of units. Thus, as in our illustrative examples, the horizontal axis may be a time scale, marked off into units of one hour each, and the vertical axis may be a distance scale, marked off into units of one mile, ox fifty miles, each. 4. In making a graph one must choose units very carefully in order to be able to get all the information on the graph, and yet make it stand out as clearly as possible. EXERCISE 26 1. The table below gives the earnings of a book agent for the latter part of July, 1915. Show the same thing graphically. Date 19 20 21 22 23 24 25 26 17 28 Earnings -$ 2.00 3.50 4.00 5.00 700 4.50 3.00 8.00 750 5.00 SUGGESTION. Represent time on the horizontal scale, and earn- ings on vertical scale. How to Represent Relationship between Quantities 51 $2.40 TIME Fig. 16. The line shows the relationship between the number of yards of cloth purchased and the total cost. 2. Figure 16 is a price graph which shows the cost of any number of yards of cloth at 12 cents per yard. From it we can find the cost of any number of yards. For example, the cost of 15 yd. is found by finding the point on the horizontal scale which stands for 15 yd., then by finding the point on the cost line directly above this point. This appears on the cost line as point A, Now, to find the cost of 15 yd. we find the point on the cost axis (O Y) horizontally opposite the point A which already stands for 15 yd. The cost proves to be 11.80. Thus we see that point A stands both for 15 yd. and for $1.80. In the same way the point B shows that 8 yd. on the horizontal scale corresponds to 96 f! on the vertical scale. The point C shows that 18 yd. on the horizontal scale corresponds to $2.26 on the vertical scale. 3. Make a formula for the cost of any number of yards of cloth at 12 ^ per yard. 52 Fundamentals of High School Mathematics Note that the graph and the formula tell exactly the same thing. The graph tells the relation between the cost and the number of yards purchased more clearly because it presents it to the eye as a picture. To tell from the graph the cost of any particular number of yards requires only a glance ; to tell from the formula or equation requires that we substitute some particular value of 11 in the equation and then that we find the value of C. 4. Draw a graph showing the price of any number of pounds of beans at 9 cents a pound. From it find the cost of 5| pounds ; of 12 pounds. 5. Now, write a formula which represents the cost of any number of pounds at 9 $ a pound. Note that the graph and the formula tell the same thing. 6. Draw a graph for the cost of a railroad ticket at 3 (f, a mile. 7. If c = .03 ;;/ is used as the equation for the cost of any railroad ticket at 3 ^ a mile, show that by letting m have particular values, such as 2, 3, 7, 10, etc., we get values for c } from which we can make the graph. 8. A number of rectangles have the same base, 5 in. Write an equation for the area of any rectangle which has a 5-inch base. (Use h for the altitude, or height.) 9. Draw a graph for the area of any rectangle whose base is 5 in. by using the equation you got in Example 8. (Let h have particular How to Represent Relationship between Quantities 53 values, such as 2, 3, 4, 7, 10, and find the cor- responding area, in each case.) 10. A west-bound train leaves Chicago at 7 a.m., going 30 miles per hour. Show graphically its progress until 4 p.m. 11. Using d = 30 t for the equation of the train in Example 10, show that the graph could have been made from the results obtained by letting t have particular values. 12. The movement of a train is described by the equation d = 25 t. Draw a graph showing the same thing. 13. A boy joined a club which charged an initiation fee of 25 cents. His dues were 10 cents each month. Draw a graph to show how much he had spent at the end of any number of months. 14. What formula or equation will represent the same thing as the graph in Example 13 ? VARIABLES AND CONSTANTS Section 22. In all the examples which you have just solved graphically there have been changing or varying quantities ; for example, in the graph of the motion of a train, the distance and the time vary as the train moves along its trip ; or in any cost graph the cost varies (that is, increases and decreases) as the member of articles varies. But in these examples, some of the quantities do not change or vary. To illustrate : the rate of the train (as in Example 10, 30 miles per hour) remains fixed, or constant, as the train moves along ; and the price per unit of any article (for example, cloth at 12 ^ per yard) remains fixed or constant in any particular example. 54 Fundamentals of High School Mathematics Thus, in any problem we may have two kinds of quanti- ties : first, those that change or vary ; and second, those that remain fixed or constant. We call them, respectively, variables and constants. For example, in the formula for the area of any rectangle whose base is 4 units, A = 4 //, it is clear that A and // are variables, and that the base, 4, remains constant. In other words, if h is 2, then A is 8 ; if h is 3, then A is 12 ; if // is 7, then A is 28, etc. Thus, h can change, but as it changes, A also changes, since A is always 4 times as large as //. Hence, 4 is the " constant " in the equation, and A and h are the " variables. " Note that there is a definite relation between A and h. A is always 4 times h. EXERCISE 27 Determine the variables and the constants in each of the following examples. Give reasons for each decision that you make. 1. c = 27rR 5. e = 10 m + 25 2. d = 4Qt 6. A=s 2 3 A== &A 1 P = 2b + 2k 4. A'=J + 4 GRAPHS SHOW THE RELATION BETWEEN TWO VARIABLES Section 23. A cost graph, such as Fig. 16, really shows the relation between the number of units (lb., doz., or yd., etc.) purchased and the total price paid. A graph of the move- ment of a train {e.g. Fig. 14) which runs at a constant rate shows the relation between the number of hours (the time) and the number of miles traveled (the distance). Saying How to Represent Relationship between Quantities 55 that these graphs show the relation between the numbers represented by them means that if we read a particular value of the time, such as 2 hr. or 5 hr., we can find the number of miles which eowesponds to that number of hours. Thus, graphs show the relation between two variables; that is, they show the values of one variable which corre- spond respectively to the values of another related variable. A formula also shows the relation or connection between the two variables. For example, the formula for the area of any rectangle with a 3-inch base, which is A — //, shows that tJie value of A must always be three times tJie value of h, or, in other words, the area is always three times the height. At first it is more difficult to understand the formula than the graph, but as you advance in mathematics the formula will become more important and significant. Thus, as we stated at the beginning of the chapter, there are three methods of showing the relationship or connec- tion between the kinds of variables we have studied: I. THE FORMULA METHOD II. THE TABULAR METHOD III. THE GRAPHIC METHOD Let us illustrate one example by each of these methods. A man walks at the rate of 6 miles per hour. Show the relation or connection between the distance he walks and the number of hours he walks. I. Formula method : D=Qh. II. Tabular method : Table 2 If the no. of hours is 1 2 S 8 10 12 then the distance is 6 12 30 4-8 eo 72 56 Fundamentals of High School Mathematics III. Graphic method : 50 ^_ j? s y^ 40 ~ ; y> _ ^** . » x^ 30 ■ S Jjr ^* v3 x »* ^^ H .?* to 20 .^ ^ x" Q ^> s ^ 10 ^ y /** J? s S 3 4 5 TIME 8 Fig. 17. The line shows relationship between time spent and distance traveled. EXERCISE 28 PRACTICE IN REPRESENTING THE RELATION BETWEEN VARIABLES Show by three methods the relation between the vari- ables in the following : 1. The area of a rectangle whose base is 8 in. and its height. 2. The cost of belonging to a club which charges an initiation fee of 50^, and 10^ per month for dues. 3. A freight train leaves Chicago at 10 a.m., at the rate of 25 miles per hour; at 1 p.m. a passenger train leaves Chicago, running in the same direc- tion, at the rate of 40 miles per hour. Show graphically at what time the passenger train will overtake the freight train. See Fig. 18 for solu- How to Represent Relationship between Quantities 57 tion. How does the graph show that one train will overtake the other? If t represents the ■ 250 " / 1 i ),' L \ 1 V^ 200 " ! yP\ ' ' ^ l i ^s \ 1 " <r' -.c^t 150 ^^ s> ^•s l/KI i 1 1 !>' / ^^ / 100 " / / * I / . s , 50 - ^ . y / ■ 1 1 ■ « s 10 12 2 3 TIME Fig. 18. The lines show relationship between the time spent and the distance traveled by each train. The point of intersection indicates the time at which they will meet and how far each travels. time of the freight train, what formula will rep- resent the distance it travels ? What will repre- sent the time the passenger train travels ? What formula will represent its distance ? 4. A slow train left Cleveland at 6 a.m., running uniformly at the rate of 30 miles per hour. At 10 a.m. a faster train left Cleveland, running in the same direction, at the rate of 40 miles per hour. Show graphically at what time the faster train will overtake the slower one. 5. A freight train left St. Louis at 7 p.m., running 30 miles per hour. At 11.30 p.m. an express train started in the same direction. Show graphically at what time it will overtake the freight train, if it runs 15 miles per hour. 58 Fundamentals of High School Mathematics Section 24. Two different kinds of graphs. We should distinguish between the two kinds of examples which we have graphed. The first kind includes all those for which no formula or equation can be made. Recall the first illustrative example in this chapter: the relation between the time of day and the temperature. Clearly, no formula can be made which will always show the relation between the two variables in this kind of example. Thus, there are only two ways to show or represent this kind of relation : (1) the tabular method, (2) the graphic method. The second kind of example which we have been graph- ing is illustrated by any of those examples for which we made a formula. For example, we have such illustrations as : the graph si towing the relation between the distance traveled by a train running at 30 miles per hour and the time the train travels. This belongs to the second kind of graph, because we can make a formula for the relation between its variables. The* formula is : d=3Ql. Thus, there are three ways to show the relation between these variables: (1) the formula or algebraic method, (2) the tabular method, and (3) the graphic method. In mathematics, we say that the second kind of graph, for which an equation can always be made, states alge- braic laws, or mathematical laws, because there is always a definite relation between the variables. The first kind of graph, for which no definite lazv or equation can be made, is sometimes called a statistical graph. It is this kind that is most frequently seen in newspapers and magazines. In mathematics, however, the other kind, that which states " laws," is nearly always used. How to Represent Relationship between Quantities 59 The next exercise will give practice in making both kinds of graphs. It is important to tell whether the in- formation to be graphed (generally called the data) can be expressed by an algebraic law or formula. EXERCISE 29 l. The following table shows the average heights of boys of different ages. Construct a graph showing this information or data. Age in years 2 4 6 8 IO 12 14 16 ia 20 Height in feet 1.6 2.6 3.0 3.5 4.0 4.8 5.2 5.5 5.6 57 3. 5. 6. Represent ages on the horizontal scale. When does the average boy grow the most rapidly ? the most slowly ? Is there an algebraic " law," or formula, which shows the relation between these two variables, age and height ? Mr. Smith joined a lodge which charged 825 initiation fee, and dues of $2 per month. Show graphically the relation between the cost of be- longing and the time one belongs. Is there an algebraic "law," or formula, which shows the relation between the variables, cost and time ? The information or data of the following table represent the area of a square of varying sides : If the side is 1 2 3 4 5 6 7 8 9 10 then the area is 1 4 9 16 25 36 49 64 81 100 oo Fundamentals of High School Mathematics Show this relation between the area of the square and its side graphically, using the verti- cal scale to measure areas and the horizontal scale to measure sides. 7. Is there an algebraic "law," or formula, which shows the relation between the variables here ? Section 25. Summary of chapter. This chapter should make clear the following truths : 1. Important facts about quantities are more easily- read and interpreted if they are represented graphically. 2. Graphs always show the relation between two varying quantities. 3. Two scales, a horizontal scale and a vertical scale, at right angles to each other, are required in order to mark off or measure the values of the varying quantities. These scales must be divided into convenient units. 4. The information or data must be tabulated in order to show it graphically. 5. There are three fundamental methods of describ- ing the relationship between related variables : a. The Formula, or Algebraic Method, of stating "Law" ; b. The Tabular Method of expressing " Law" ; c. The Graphic Method of expressing " Law." EXERCISE 30 l. If n represents a boy's present age, state in words what the expression n + 7 = 22 means. How to Represent Relationship between Quantities 61 2. Give a formula for the base of a rectangle when the area and height are known. 3. Represent the number of cubic yards in a box- shaped excavation when the dimensions are ex- pressed in feet. 4. If m, s, and d represent the minuend, subtra- hend, and difference respectively, what formula will show the relation between these numbers? 5. Show by a formula the relation between the product,/, multiplicand, M, and multiplier, vi. 6. Give the meaning of the formula i=prt. 7. Divide each side of the formula V= Iwh by Iw and tell what the resulting formula means. 8. Give a formula for the volume of a cube whose edge is s. 9. Evaluate the above formula when s = 3.2. 10. Translate into words the formula d = rt. 11. Divide each side of the formula d— rt hy r and tell what the resulting formula means, 12. In the formula c = np, n represents the num- ber of articles bought, p represents the price of each, and c represents the total cost. Trans- late it into a word statement. 13. Divide each side of the formula c = np by n, and tell what the resulting formula means. 14. Does x = 4t satisfy the equation x 1 + 6 x = 40 ? 15. Solve the equation 10 j/ + 7 = 52 + 4 y. 16. Is the equation x +y + 3 = 20 satisfied if x = 8 and y = 9 ? Can you find any other values of x and y which will satisfy this equation ? 02 Fundamentals of High School Mathematics (*) 17. Solve each of the following equations, thinking of each example as asking a question : 2^ = 5.5 y (b) .5/ =17 (/) 1.5* = 45 (<:) .4/ = 80 Qr) f 2 = 60 34 ,„r ... ' h {a) ~- = 2 x V) 4.L>f> (//) .34 = 85 Tell what you do to each side of the equation ; that is, tell whether you add, subtract, multiply, or divide, on each side. REVIEW EXERCISE 31 1. Write in algebraic language : The volume of a sphere is four thirds the cube of the radius times 7r. 2. The weights of a baby boy who weighed 8 lb. at birth are given for each month of his first year by the table : Month 1 2 3 4 5 6 7 8. 3 io 11 12 Weight 9* 11* I2f 14* 151 I65 18 19 19| 20 21 22 Represent this graphically. 3. Which would you rather have, ix + by dollars, or 5 x 4- 4 y dollars, if x = % 20 and y = $ 1 6 ? 4. If V is the volume of a cone, b the area of its base, and // its height, then V= \bh. Write this formula in words. How to Represent Relationship between Quantities 63 5. Nurses keep temperature records of fever pa- tients. For one patient the following degrees of fever were noted : 2, 5.4, 4, 6.1, 4.5, 6.5, 5.3, 6.7, 4.5, 6,5, 5.9, 6.2, 6.9, 5, 6.4, 4.7, 5.8, 7.6. These readings were taken an hour apart. Show graphically this patient's successive tem- peratures. 6. If you know that 10a - 7 = ±a + 35, then what do you do to each side to get 10^ = 4^ + 42? How do you get 6 a = 42 ? Why does a = 7 ? Prove that a = 7. 7. Get the hourly temperature for 24 hours from your daily paper, and construct a graph to rep- resent the changes in temperature. 8. The sum of two numbers is 18 and their differ- ence is 4. What are the numbers ? 9. Show that 2 a and a 2 are unequal by choosing some particular value for a, such as a = 6. Do you think there is any possible value for a which would make 2a = a 2 ? 10. What product is obtained by using. 7 as a factor twice ? by using 2 a as a factor three times ? CHAPTER V HOW TO FIND UNKNOWN DISTANCES BY MEANS OF SCALE DRAWINGS: THE FIRST METHOD Section 26. We need to know how to find unknown dis- tances. The methods of mathematics are really all planned to help us find unknown values. The equation, which we have studied so carefully, is the best algebraic tool with which to do that. Many times, however, in practical life work the unknown values that we need to know are dis- tances. For example, the surveyor may need to know the distance across a river and may not be able actually to measure it. Or, he may need to know the distance be- tween two points, with some other intervening object between which prevents him from measuring it directly. Now, mathematics has given us three ways to find such an unknown distance. In Chapters V, VI, and VII we shall discuss these methods. The first method is to make a scale drawing, which will include in some way the unknown distance. Next, there- fore, we shall study how to determine unknown distances by means of scale drawings. Before we take up that particular subject, however, we must study. how to measure the lines and angles which make up scale drawings. Section 27. The measurement of lines. We are already familiar with certain methods of measuring distances. For example, we have measured the length of lines, such as the distance from A to B or from C to D. If we use a metric scale, in which the units are centimeters, the distance from A L C D^ 6 4 Finding Unknown Distances by Scale Drawings 65 A to B y which is read "line AB," is 5.08 centimeters long, and the line CD is 6.35 centimeters long. If we use a foot- rule in which the units are inches, the distance between A and B y or the line AB, is 2 inches, and line CD is 2.5 inches. Note here that the distances or lengths that we obtain for these lines depend upon the kind of scale, or kind of unity that is used in measuring. THE MEASUREMENT OF ANGLES Section 28. An angle is determined by one line turning about another. In order to construct scale drawings, we must know how to measure angles. Let us think of an angle as being formed by one line turning, or rotating about a fixed point on some fixed or stationary line. The line O Y turns or rotates about point O. For example, in Fig. 19 Fig. 19, think of AX as a fixed, or stationary, line. (It is easiest always to take this line as horizontal.) Think also of another line, say O Y y as turning, or rotating, about some point on the fixed line AX 9 say point O. As the line OY rotates about the point O f it constantly forms a larger and larger angle with the fixed line AX. (The symbol for angle is Z.) The point O, about which the line turns, is always the point at which the two sides of the angle meet, and is called the vertex of the angle. 66 Fundamentals of High School Mathematics The arrow is drawn to indicate that the line OY is turn- ing, or rotating, about the point 0. Section 29. The unit of angular measurement. Just as we have units and scales for measuring straight lines, so we have units and scales for measuring angles. Evidently the unit with which we must measure the size of the angle is one that will measure the amount that the line has rotated about the fixed point. Figure 20 shows that we can think of the rotating line as turning clear around until it occupies its original position again. That is, any point P on the line OX has turned through a complete circle in rotating about O and returning to its original posi- tion. This suggests that the unit with i • i Fig. 20 which we measure angles zvill be some definite fraction of the circle. Fdr a long time people have agreed that the circle be divided into 360 units and that each one of these units of angular measure be called a degree. The symbol used for degree is a small ° placed at the right above the number. For example: 45° is read "45 degrees." Thus, Figs. 21, 22, and 23 illustrate angles of different sizes or of differ- ent numbers of degrees. Finding Unknown Distances by Scale Drawings 67 Fig. 21 Fig. 22 s 135 Fig. 23 Section 30. The PROTRACTOR: How to measure angles. Just as we use foot rules, yardsticks, meter sticks, etc., to measure straight-line distances, so we have an instrument called a protractor to measure angular distances. Figure 24 shows that the circular edge of the Fig. 24. A protractor for constructing and measuring angles. 6S Fundamentals of High School Mathematics protractor is marked off (*.*■ is " graduated ") into degrees. Note from the figure that the protractor is. divided into 180 equal parts (half of the total number of angular units in the circle), called degrees. Sometimes the whole circle is used and marked off to give 300°. Fig. 25 The next figure, Fig. 25, shows how to measure an angle with a protractor. First, lay the straight edge of the pro- tractor so that it will fall exactly upon one of the two lines that form the angle, and with the center of the pro- tractor exactly upon the vertex, O, of the angle. Then the other side of the angle, OB y for example, will appear to cut across the circular edge of the protractor. Now count the number of degrees from the point where the curved edge of the protractor touches OA to the point where it crosses the line OB. Hence, in Fig. 25, the angle A OB contains 54°. It is very important for us to be able to read angles accurately. The next exercise will give you practice in reading angles. Finding Unknown Distances by Scale Drawings 69 EXERCISE 32 PRACTICE IN MEASURING ANGLES 1. Measure each of these angles with a protractor. B Fig. 26 Fig. 27 Fig. 28 Compare angle A and angle C. Which has the longer sides ? What effect has the length of a side of an angle upon the size of the angle ? Measure each angle of triangle ABC. From the results of your measurement, what is the sum of all three angles of this triangle ? Fig. 29 jo Fundamentals of High School Mathematics How large is Z.r? How many de- i n Z y ? ? How X g r e e s in Z; many degrees Fig. 30 in the sum of the angles of this tri- angle, XYZ? 5. Draw with the protractor an angle of 30° ; 45° ; 60°; 100°. 6. At each end of a line 6 cm. long draw angles of 50°. Produce these lines until they meet, and measure the angle formed by them. How many degrees in it? Compare the lengths of the lines you drew. How many degrees does the sum of the three angles of this triangle make ? 7. Draw a triangle such as triangle ABC, so that AB = 4 inches, angle ^ = 60° and AC = 3 inches. Then find the number , ._. of degrees in an- ■**■ )<- gle B and angle C 8. Construct triangle ABC so that AC=r> cm., angle C=40°, and CB = 5 cm. Compare angle A with angle B. How many de- grees in each ? Explanation : A triangle having two sides equal, such as AC and CB, is an isosceles triangle. It is proved in geom- Fig. 32 Fig. 31 Finding Unknown Distances by Scale Drawings 71 etry that the angles opposite these equal sides are always equal; for example, angle A = angle B. How many degrees ought there to be in either angle A or angle B ? Section 31. How to describe an angle. An angle is described by using three letters, i.e. the letter which repre- sents the vertex is written between the two letters at the ends of the sides. Thus, Z 1, in Fig. 33, is read as angle A OB or angle BOA, and is written Z A OB or Z BOA. In the same way, Z 2 is read angle BOC or angle COB, and is written Z BOC or Z COB. Fig. 33 EXERCISE 33 PRACTICE IN READING ANGLES 1. Why would it not be clear to read Z2 as Z O? 2. Read the angle formed by lines OA and OB. 3. Read the angle formed by lines OB and OC. 4. Determine the number of degrees in ZAOB, in Fig. 34, without using the protractor* Fig. 35 72 Fundamentals of High School Mathematics 5. If in Fig. 35 you know that angle ABC is 40° and that Z BCA is ( <>0°, could you find Z CAB without measuring it? How ? How large is it ? Section 32. We must be able to find unknown distances which cannot be measured directly. The preceding sec- tion took up only examples in which the distances, linear and angular, could be measured directly, by means of instruments. There are many instances, however, in which the lengths of the lines and the sizes of the angles cannot be measured directly. For example, consider the case of finding the distance across a river, or the height of a tree, which we mentioned at the beginning of the chapter. In cases like this we need indirect methods of measuring. Mathematics makes it possible for us to determine the lengths of such lines by measuring the lengths of other lines and the sizes of angles that are related to them. This leads us to the main topic of this chapter. HOW TO FIND UNKNOWN DISTANCES BY MEANS OF SCALE DRAWINGS Section 33. How to draw distances to scale. One of the methods that you will use commonly in indirect measure- ment is that of drawing distances "to scale." So much use is made of mechanical drawings that we need to be very proficient in making them and in reading them. Let us take a simple illustration of the drawing of distances "to scale " and of measuring distances on scale drawings. Finding Unknown Distances by Scale Drawings 73 Illustrative example. A man starts at a given point and walks 2.5 miles east, then 2.5 miles north. How far is he from his starting point ? First, set point 0, in Fig. 36, as his starting point. East is meas- ured to the right of point O and north above point O. Second, to represent distances "to scale," we need to select a b\ c •ft x? >/ <- y 6 , n c % in x! ^ (N <^ ^ ,rj xf V *£3 «? ^ , ll § 5) 1 •*/ f k Ej as t . 2. 5i iii let — > F > Fig. 36 wniY 0/ distance on the scale which will represent a unit of distance in the example. Let us take, for example, three quarters of an inch on the drawing to represent each mile which the man actually walks. This is indicated on the scale drawing (Fig. 36) by writing "Scale = § in. to 1 mi." It is very important to select the scale unit carefully and always to indicate the scale that has been used on the drawing. Third, to represent the man's path, we lay off OB horizontally to the right of O, 2.5 miles (on the drawing this amounts to If inches) and BC vertically, 2.25 miles. Then, by using the cross-section paper as a scale, we can measure at once the distance, OC, that the man is from his starting point. The distance is 2.63 inches on the figure, or 3.54 miles actually. 74 Fundamentals of High School Mathematics This work illustrates by a very simple example how we make scale drawings. Mechanical drawings made " to scale " are used very commonly by such workers as archi- tects, carpenters, machinists, and engineers. EXERCISE 34 PRACTICE IN FINDING UNKNOWN DISTANCES BY THE CONSTRUCTION OF SCALE DRAWINGS 1. Draw to the scale 1 cm. to 2 ft. a floor plan of a room 28 ft. by 20 ft. By measuring the distance diagonally across the plan, compute the diagonal of the room. 2. Draw a plan of a baseball diamond 90 ft. square and find the distance from first base to third base. Use 1 cm. to represent 20 ft. 3. Two bicyclists start from the same point. One rides 12 miles north and then 8 miles east; the other rides 10 miles south and then 6 miles west. How far apart will they be ? Use the scale 1 cm. to 2 mi. 4. Draw to the scale 1 cm. to -4 ft. a plan of the end of a garage such as in Fig. 37. Find the height from the floor to the top of the roof. Fig, 37 Finding Unknown Distances by Scale Drawings 75 5. In Fig. 38 AB is 200 ft., angle A is 60°, and angle B is 50°. Draw to the scale 1 cm. to 50 ft. and determine the length of AC and BC. 6. A surveyor sometimes finds it necessary to measure the distance across a swamp, such as Fig. 38 Fig. 39 AB in Fig. 39. He measures from a stake i4toa stake C, 120 ft. From C to B he finds it is 100 ft. Find, by a scale drawing, the dis- tance AB across the swamp, if angle C is 85°. How could a surveyor find the distance from A to B, if there were some obstacle in the way j6 Fundamentals of High School Mathematics «^> A Fig. 40 B preventing his measuring directly the distance AB? 8. Find by a scale drawing the distance AC across Fig. 41 the river, if it is known that angle A = 80°, AB = 200 ft., and angle B = 70°. 9. Illustrative example. A boy wishes to determine the height of a flagpole. A scale drawing will aid him in j l q r; \ s \ O \ 6 P ^ \ known L f 1 \ ! , B l \ known Z.\ rp <— —80' Known A Fig. 42 Finding Unknown Distances by Scale Drawings 77 doing this. For example, he can measure a line of any length on the ground out from the base of the flagpole. Suppose he takes a line 80 feet long. Then he sets at E an instrument called a transit, with which he can read the angle between the horizontal base line and the line of sight from E, where he stands, to H, the top of the pole. He knows also that the angle B is a right angle. So he knows the length of the line EB, the size of the angle E and the angle B. He constructs a scale drawing to represent the known lengths and the known angles. From this drawing he is able to " scale " or measure the height of the flagpole. The angle BEH or angle E between the horizontal and the line of sight in this example is called the angle of elevation. If the boy had taken a longer base line, what would have been true of the size of the angle of elevation with respect to what it was before ? 10. If the angle of elevation in Fig. 42 is 40° when the observer is 80 ft. from the foot of the pole, find its height. 11. A flagpole 50 ft. high casts a shadow 60 ft. long on level ground. What is the angle of elevation of the sun ? If the length of a shadow cast by this pole increases, what conclusion can be drawn concerning the angle of elevation ? 12. The angle of elevation of the top of a tree is 42° when the observer stands 30 yd. from the tree. How high is the tree ? If the distance from the observer to the tree decreases, what change in the angle of elevation follows ? 78 Fundamentals of High School Mathematics 13. On the top of a church is a tall spire. At a point P on level ground 75 ft. from the point D directly beneath the spire, the angle of eleva- tion of the top of the spire is 38°. At a point Q 150 ft. from D, in line with P and D, the angle of elevation of the top of the spire is 28°. Find the distance from Q to the top of the spire. 14. An anchored balloonist from a height HT, Fig. 43, of 2500 yd., observes the enemy at D. BALLOON Fig. 43 He wishes to compute the distance DT, on level ground. To do so, he measures the angle which is formed by the horizontal line HL and the line of sight HD. This angle is called the angle of depression, and has the same num- ber of degrees as the angle of elevation. (Can you see that this is true ?) He next finds angle THD by subtracting the angle of depression from 90°, angle THL. He then knows enough Finding Unknown Distances by Scale Drawings 79 about the triangle DTH to make a scale draw- ing of it. Find DT if the angle of depression is 35°. 15. From the top of a lighthouse 80 ft. high the angle of depression of a ship is 35°. How far is the ship from the base of the lighthouse? Compare your result with that obtained by the other members of the class. 16. From the top of a cliff 120 ft. above the sur- face of the water the angle of depression of a boat is 20°. How far is it from the top of the cliff to the boat ? 17. A searchlight on the top of a building is 180 ft. above the street level. Through how many degrees from the horizontal must its beam of light be depressed so that it may shine directly on an object 400 ft. down the street from the base of the building ? How does your result compare with that obtained by other pupils ? 18. From the tenth story of a building the angle of depression of an object on the street level is 30° ; from the eighteenth story the angle of de- pression of the same object is 50°. Find the distance from the object to the base of the build- ing if the second observation point is 80 ft. above the first observation point. 19. An observer is 200 ft. from the ground. The angle of depression of a point A is 24°, of a point B 42°, and of a point C 15°. Which point is closest to the observer ? farthest from the observer ? 8o Fundamentals of High School Mathematics 20. In Fig. 44 measure (1) the angle of elevation of point D from point E ; (2) the angle of de- pression of point E from point D. Compare these angles. Note that line DH is parallel to EC, and that these angles are formed by line DE cut- ting (or intersecting) these two parallels. In geometry it is proved that such angles are al- ways equal. 21. Is the scale drawing a very accurate method of determining unknown distances ? Do many of the pupils in the class get the same answer for any particular example ? Why ? A NEW WAY TO COMPARE TWO QUANTITIES; NAMELY, TO FIND THEIR RATIO Section 34. We have been comparing two quantities by finding how much larger or smaller one quantity was than another quantity. For example, if the line AB is 4 units long and the line CD is 6 units long, we should have said, in describing the comparative lengths, that the line AB was 2 units shorter than line CD, or that line CD was 2 units longer than the line AB. A B C D Another method, however, of comparing quantities is used very extensively in mathematics. It is the method of dividing one quantity by the other, or finding the quotient Finding Unknown Distances by Scale Drawings 81 of the two quantities. Thus, to compare the line AB with the line CD we divide AB by CD, which gives : AB = 4: CD 6* This result is read " the quotient, or ratio, of AB to CD is equal to four sixths." This process is described as find- ing the ratio of the two lines. The ratio of two numbers means, then, the quotient which results from dividing one of the numbers by the other. Thus, the ratio of 5 to 10 is |-, and is written T 5 ^ = \. The ratio of 10 to 5 is 2, and is written -5°-= 2. In the same way the ratio of 1 in. to 1 ft. is -^ ; the ratio of \ to | is f ; and the ratio of ^x to 1 x is 4:X 4 7 x 7 EXERCISE 35 PRACTICE IN DEALING WITH RATIOS OF NUMBERS 1. What is the ratio (in lowest terms) of 10 to 12 ? of 20 to 24 ? of 15 to 8 ? of 25 to 30 ? Show that 5x and Qx represent all pairs of numbers whose ratio is f . 2. Give several pairs of numbers having the ratio f . Show that 3x and 4.r represent all pairs of num- bers having this ratio. 3. The ratio of two numbers, a and b, is \ . What is b when a is 40 ? 4. The ratio of two lines, m and ;z, m |. Find the length of m if 71 n is 18 in. 5. Divide 40 into two parts whose ratio is f . 6. A father and his son agreed to divide the profits from their garden in the ratio of |. Find each one's share if the total profits were $210. 8j Fundamentals of High School Mathematics 7. The ratio of Z A to Z. B is f. Find Z B when ZAis 80°. Fig. 45 8. The ratio of the areas of the two squares, S x and S 2f is \. Find the area of each if the sum of their areas is 45 sq. in. 10. Fig. 46 Divide an angle of 90° into two angles having the ratio of 4 to 5. Measure each angle, Fig. 47, with a protractor. Find their ratio. Fig. 47 Finding Unknown Distances by Scale Drawings 83 11. In the two triangles ABC and XYZ, what is the ratio of Z CtoZZ? Of ZAto ZX? Oi Z B to Z Y? Do these triangles have the same shape ? Do all triangles have the same shape ? Fig. 48 12. Find the ratio of two lines if one is 2 feet long and the other 3 yards long. 13. What is the ratio of 3 pints to 4 quarts ? 14. 2 = A. 15 . 4.5. 16 . 5 10. n 15 3 6 In 17. Find the ratio of AB to CD, Fig. 49, by measur- ing the length of each line. Express the result decimally. A B Fig. 49 18. A school baseball team won 7 of the 10 games it played. What was its standing in percentage. ? Solution : The standing of the team in percentage means the ratio of the number of games won to the number played, expressed decimally. Thus, the standing of this team is ex- pressed by the result obtained from changing the ratio, T 7 o, to a decimal. This gives .70 or .700. 84 Fundamentals of High School Mathematics 19. What was the percentage or standing of a team which won 9 of its 12 games ? 20. The winning team in one league won 14 of its 18 games ; and the winning team in another league won 15 of the 19 games it played. Which had the higher percentage ? 21. Express each of the following ratios as a deci- mal, correct to two places : / \ 3 «>1 ( \ 5 {e) 16 ( ^| ... 22 W 81 «.£ C/)f (/ ' } 45 , ., 16.2 {J) 21 22. If 12 quarts of water are added to 25 gallons of alcohol, what is the ratio of the water to the entire mixture ? Express decimally. REVIEW EXERCISE 36 l. A boy knows that AB is 100 ft. and that Z. B = 40°. From this information can he con- struct a scale drawing for the triangle ? Give reasons for your answer. 100 ft 2. What facts or data must be known about a tri- angle before you can make a scale drawing of it ? Finding Unknown Distances by Scale Drawings 85 3. How does the scale of a drawing illustrate ratio ? 4. A tree 90 ft. high casts a shadow 140 ft. long. Find from a scale drawing the angle of elevation of the sun. 5. The following table shows the humber of feet required to stop an automobile running at vari- ous speeds. At a speed of (miles per hour) IO ]5 20 25 30 35 40 50 a car should stop (feet) 9.2 20.8 37 5Q. 83.3 104. 148. 231. Represent this graphically, measuring the speed on the horizontal axis. 6. Represent in the briefest way the product of five x's ; the sum of five x's. 7. Does a 2 b = ab 2 if a is 4 and b is 3 ? 8. Can you construct a triangle similar to Fig. 50 if you know that AB is 200 ft. and AB = 40°? Why? 9. What must be known about a triangle before you can construct it accurately ? CHAPTER VI A SECOND METHOD OF FINDING UNKNOWN DISTANCES: THE USE OF SIMILAR TRIANGLES Section 35. Scale drawings are very inaccurate. In the last chapter we saw that scale drawings could be used to find unknown distances, either linear or angular. The results obtained, however, were very inaccurate. Seldom did many of you get the same answer for any example. Therefore we need more accurate metliods for determining unknown lines and unknown angles. This chapter, and the next one, will show methods that depend less upon the accuracy or skill of the person who "scales" or measures. The first method, based ?tpon geometrical figures of exactly the same shape, will be explained now. Section 36. What are similar figures? You have al- ready seen many objects or figures of exactly the same shape. A scale drawing has the same shape as the figure from which it was made ; on a photographic plate the figure is the same in outline or shape as the original ; the map of a state has the same shape or outline as the state itself. Figures which have the same shape are said to be simi- lar figures. Which of the following figures are similar in shape ? o Fig. 53 Fig. 54 Fig. 55 Fig. 51 Fig. 52 Fig. 56 86 Finding Unknown Distances by Similar Triangles 87 Similarity in shape in geometrical figures is a very im- portant principle that we are able to use in many ways in mathematics. Before this can be taken up, however, we need to be perfectly clear as to what is meant by similarity of figures. The next exercise will help to do this. EXERCISE 37 PRACTICE WITH SIMILAR TRIANGLES 1. Draw a line XY twice as long as AB, Fig. 57. At X draw an angle equal to angle A. At Y draw an angle equal to angle B. Produce the sides of these angles until they meet at Z. Measure the angle formed by these sides. How should it compare with angle C? Why? 2. (a) Angle A corre- sponds to what angle in your triangle ? (b) Angle B corresponds to what angle in your triangle ? {c) What is true, then, about the corresponding angles of the two triangles ? 3. Measure the side in your triangle which corre- sponds to AC, and the side which corresponds to BC. What is the ratio of AB to XY, or what is the value of ? of ? of ? What does X Y X/l, Y Z, this tell about the ratios of corresponding sides ? Fig. 57 88 Fundamentals of High School Mathematics Construct a tri- angle larger than Fig. 58 but having its angles equal to the angles of Fig. 58. Is your tri- angle the same shape as Fig. 57 ? 1.5" Fig. 68 After careful measurement find the ratio of AB to its corresponding side in your triangle. Then find the ratio of AC to its corresponding side. Compare these ratios. Fig. 59 Fig. 60 5. In Figs. 59 and 60 it is known that AC 2 3' BC 2 , AB 2 XZ In other words, it is known that the ratios of corresponding sides are equal. Measure each angle of each triangle and compare each angle in Fig. 59 with its corre- sponding angle in Fig. GO. State carefully the conclusion to be drawn from this comparison. Are these triangles of the same shape ? Section 37. Similar triangles. The previous exercises illustrate two very important and widely used truths about similar triangles : Finding Unknown Distances by Similar Triangles 89 (1) IF IT IS KNOWN THAT THE ANGLES OF ONE TRIANGLE ARE EQUAL RESPECTIVELY TO THE ANGLES OF ANOTHER TRIANGLE, IT FOLLOWS THAT THE RATIOS OF THE CORRESPONDING SIDES OF THE TRIANGLES ARE EQUAL. (2) IF IT IS KNOWN THAT THE RATIOS OF THE CORRESPONDING SIDES OF TWO TRIANGLES ARE EQUAL, IT FOLLOWS THAT THE ANGLES OF ONE TRIANGLE ARE EQUAL RESPECTIVELY TO THE ANGLES OF THE OTHER TRIANGLE. Thus, to state that two triangles are similar is equiva- lent to stating that the corresponding angles are equal \ and that the ratios of the corresponding sides are equal. This principle or truth is used very much in mathe- matics. To illustrate, suppose we know that the angles of one triangle are equal respectively to the angles of another triangle ; then we also know that the ratios of correspond- ing sides are equal. Hence, we can make an equation from these equal ratios and from this equation find important unknown distances. The next exercise will show how this is applied to finding the length of lines. EXERCISE 38 ADDITIONAL PRACTICE WITH SIMILAR FIGURES 1. Figure 61 is a right triangle. Why? If angle A is 30°, find an- gle C. If angle A is one fourth of angle C, find the size of each angle. 2. In a right triangle one of the acute angles (that is, one of the angles smaller than a right angle) is 40°. Find the other acute angle. Fig. 61 90 Fundamentals of High School Mathematics Fig. 62 Fig. (>3 3. 4. 6. Figures 62 and 63 are right triangles. If angle A = angle D, are the triangles similar ? Why ? In Figs. 62 and 63, if EF=6, ED = 8, and CB=3, what must AB equal? To solve this problem we use the principle that the ratios of the corresponding sides of similar triangles x 3 equation q=77- 8 D are equal. This gives the What, therefore, is AB ? The sides of a triangular plat of ground are 150 ft, 100 ft, and 125 ft, respectively. The side of a scale drawing of this plat, correspond- ing to the 150-foot side, is 5 cm. Find the side of the scale drawing corresponding to the 100-foot side. Solve as in Example 4. The sides of a triangle are 3, 4, and 5 cm. The shortest side of a similar triangle is 16 cm. Find the other sides of the second triangle. A house is 36 ft. high and the garage is 16 ft. high. If the house is represented in a drawing as 18 in. high, how high should the drawing Finding Unknown Distances by Similar Triangles 91 of the garage be ? What mathematical prin- ciple is used to show this ? 8. Two rectangular gardens are the same shape, but of different size. The larger one is 72 ft. by 84 ft. If the length of the smaller one is 40 ft., what must be its width ? 9. Two angles of one triangle are equal respec- tively to two angles of another triangle. Are the triangles similar ? Why ? 10. Line AB is parallel to line CD. Would they meet if produced, either to the right or to the left of the third line MM? Meas- ure Z 1 and Z 2. These angles are called cor- }sj' responding fig. 64 a angles of parallel lines. 11. In Fig. 64 measure the other pair of correspond- ing angles, Z 3 and Z 4. What do you find ? These two exercises illustrate a very important fact in mathematics ; namely, that the corresponding angles of parallel lines are always equal. Later on this will be proved without measuring the angles ; that is, without any possibility of error. You will make use of this fact with- out again measuring the angles. £ 12. In triangle ABC, DE is drawn />v parallel to AB. DoesZl = Z2? 1^4- -_XE Why ? Is triangle DEC similar j\^ -^B to triangle ABC? Why ? fig. 64 b 92 Fundamentals of High School Mathematics 13. 14. 15. 16. In Example 12, DC= 12, AC = 21, and CE = 10. Show how BC can be found, by using the prin- ciple that the ratios of corresponding sides of similar triangles are equal. What is the length of BC? A boy wishes to measure the height of a tree. He notes that the tree, AC, its shadow, AB y c J 2« STICK k-10'— I Fig. 65 and the sun's ray, CB, passing over the top of the tree, form a triangle. He measures the shadow and finds it 100 ft. long. At the same time a vertical stick 4 ft. high makes a shadow 10 ft. long. Why is the triangle formed by the stick, its shadow, and the sun's ray passing over the top of the stick similar to the other triangle ? How can the boy find the height of the tree from the similar triangles ? What is its height ? A Boy Scout wagered he could find the distance between two trees, A and B, on opposite sides of a river, without crossing it. Could he do it, and if so, how ? If not, why not ? A crude way to measure the height of an ob- ject is by means of a mirror. Place a mirror Finding Unknown Distances by Similar Triangles 93 horizontally on the ground at M y and stand at the point at which the image of the top of the Fig. 66 object is just visible in the mirror. Show how, by measuring certain distances, this would enable one to compute the height of the object. 17. In triangle ABC, DE is parallel to CB. Show that triangle BED is similar to triangle ABC. If BC= 10, ED = 5, and AE = 8, what is AB ? 18. Show that in Fig. 67 the ratio of DE to AE, or DE AE , remains the same even if DE is drawn in different positions (always parallel to CB). n O \v. D M r V _L B E * ? - « £ ^ Fig. 67 94 20. Fundamentals of Higli School Mathematics % Y 19. Fig. (i8 Figure 08 shows two triangles, with the size of each angle indicated, which a teacher drew upon the blackboard for an examination. She asked the following questions about the two triangles : (a) Are they similar triangles ? AB _AC ZY AC Why? {b) Does (c) Does (d) Does XY ZY_ AB~~ XY _BC ? ' ZY' XZ , : BC Why? Why? Why? (<r) Does the ratio of any two sides equal the ratio of any other two sides ? How would you have answered these questions ? The sides of a small triangle are 3, 4, and 6. Is it similar to a larger triangle whose sides are 15, 18, and 80 ? See Section 37. SUMMARY OF THE IMPORTANT POINTS OF THE CHAPTER It is important to have clearly in mind the following im- portant conclusions from the chapter : Finding Unknown Distances by Similar Triangles 95 1. If you know that the angles of one triangle are equal respectively to the angles of another tri- angle, then you know that the ratios of the corre- sponding sides are equal In other words, you can make an equation, and thereby find an unknown side. 2. If you know that the ratios of the corresponding sides of two triangles are equal, then you know that the angles of one triangle are equal respec- tively to the aggies of the other triangle. 3. The corresponding angles of parallel lines are equal. 4. Unknown distances may be found by means of similar triangles. REVIEW EXERCISE 39 1. Translate into words : \y + 3 = y + 21. 2. If A, B y and C represent the number of degrees in the respective angles of a triangle, we know that A + B + C = 180°. Why ? What is A if £ = 40 D and C=65°? 3. If five times a certain number is divided by 2.7, the result is 3. What is the number ? 4. Given the formula V= lwh> find a formula for /; for w. 5. A boy receives C cents an hour for regular work, and pay for time and a half when he works over- time. What will represent his earnings for 6 hr. overtime? Evaluate this when £7=50. 96 Fundamentals of High School Mathematics 6. The number of years that a man at various ages may expect to live, as determined by insurance experts, is as follows : If a man is (age in. years) 10 15 20 25 30 35 40 45 50 he may still live (years) 49.6 45.2 41. 37 33.1 29.2 25 6 22.2 18.9 Construct this graphically, representing ages on the horizontal axis. 7. From the graph, find how much longer a man 21 years old may expect to live ; a man 32 years old. CHAPTER VII HOW TO FIND UNKNOWNS BY MEANS OF THE RATIOS OF THE SIDES OF THE RIGHT TRIANGLE Section 38. The advantage of the RIGHT TRIANGLE in finding unknowns. It should be clear by this time that mathematics gives us methods of finding unknown quanti- ties. The equation is the most important tool for doing this, for the reason that when we solve a problem we have to make an equation. This equation must contain the un- known quantity together with other known quantities which are related to it in some way. In the last chapter we saw that an equation could be formed from the ratios of corresponding sides of similar triangles and that by that means we could find an un- known length. Two facts, however, make that method less satisfactory than the one we shall study in this chapter: (1) we must always be certain the triangles are similar, or we have no right to make an equation, and (2) the method is cumbersome because we must always use two triangles. There is a particular kind of triangle whose properties can be used to find unknown distances accurately and at the same time more easily than by any other method. It is the right triangle. The most important fact about the right triangle is found in connection with the ratios of its different sides. I. THE TANGENT OF AN ANGLE Section 39. The ratio of the " side opposite " a given angle to the " side adjacent " the given angle, i.e. the TANGENT of the angle. You will recall that in a right tri- angle one angle is 90° and the sum of the two acute angles equals 90°. (Why?) In finding unknown distances by 97 gS Fundamentals of High School Mathematics means of right triangles we shall always deal especially with one of the acute angles. Therefore, in referring to A side adjacent to 30 °L ^ Fig. 69 ° hypotenuse Fig. 70 the sides of a right triangle, when dealing with a given angle, we shall speak of them as they are described in Figs. 69 and 70. If angle B is the acute angle with which we are concerned, then side AC is the " side opposite" Z B, and side AB is the " side adjacent " Z B. The side opposite the 90° angle is always called the hypotenuse. Some exercises will show the importance of the r&tio the "s ide o pposite " the "side adjacent" an acute angle of a right triangle. Finding Unknowns by Ratios of Sides of Triangle 99 EXERCISE 40 SOME EXPERIMENTS TO FIND THE NUMERICAL VALUE OF THE RATIO OF THE ■ ■ SIDE OPPOSITE ' » TO THE ' ■ SIDE ADJACENT " A 3CP ANGLE OF A RIGHT TRIANGLE T Fig. 71 In Fig. 71, a t is the " side opposite " the 30° angle and b 1 is the "side adjacent" the 30° angle. Measure a x and b v Now find the numerical value of the ratio of a x to b x by dividing the length of a x by the length of b v Record your results in Table 3. In Fig. 72, a x and b x are respectively the " side opposite" and the "side adjacent" an acute angle of 30°. Measure each and compute the ratio ~r to b \ results in Table 3. two decimal places. Record your Fig. 72 ioo Fundamentals of High School Mathematics 3. Draw any other triangle similar to those above, but with much larger sides. Measure the "side opposite " and the " side adjacent" the 30° angle and compute their ratio as before. Record re- sults, as before, in Table 3. Table 3. Record here the results of measuring the sides of right triangles and of computing the ratio of the " side opposite " to the " side adja- cent " an acute angle 0/"3O°. Table 3 Length of a a Length of b x Ratio of a^ob^e.,-^) Fig. % fig- Fig. What do you notice in the table about the nu- merical values of the ratios the "side opposite" an acute angle of 30° a 1? the "side adjacent" an acute angle of 30° b x The members of the class should compare re- sults, to see what result seems most likely to be the true one. If great care is taken in measur- ing, the ratio should be very close to .58 in each triangle. Why should it be the same in each triangle ? Finding Unknowns by Ratios of Sides of Triangle 101 4. In Fig. 73, CB, DE, and GF are perpendicular to AB. Is tri- angle AFG V. similar to tri- D^^ angle AED1 G^ Why? Is ^*^ i either of the s^ i smaller tri- K s^ i angles sim- F E I ilar to the Fig. 73 large triangle ? Why ? From this, why does the GF ratio of GF to AF, or — -, equal the ratio of DE AF DE to AE, or ? If you measured these lines. AE J ■ and computed the ratios, what would you expect to be true of the results ? Section 40. This last example is very important, because GF it shows, without measurement, that the ratio equals AF the ratio BE AE But this is the same as saying that the ratio of the "side opposite " to the "side adjacent" a 30° angle in one right triangle is ALWAYS equal to the ratio of the "side opposite" to the "side adjacent " a 30° angle in any other right triangle. The length of the sides may be far different, but the ratios should be the same. This shows that the ratios obtained in the table should have been the same, if it were possible to draw and measure without error. We shall now make use of the fact that the numerical value of the ratio of the "side opposite" to the "side ad- 102 Fundamentals of High School Mathematics jaccnt" an acute angle of 30° (in a right triangle) is ap- proximately .58, EXERCISE 41 l. A man wishes to determine the height of a smokestack. He finds that the angle of eleva- tion of the top of the smokestack, from a point 200 ft. from the base of the smokestack, is 30°. Fig. 74 Solution : = .58. 200 h = 200 x .58. h = 116 ft. (Why ?) (Why ?) Note here that — is the ratio of the " side opposite " to 200 ** the "side adjacent" the 30° angle. From previous work we know that this ratio is .58. Thus, we can make the equation =.58. H 200 2. In triangle ABC, angle A is 30° and angle C is 60°. Find CB if AB is 75 yards. What is CB if AB is 10 inches ? Draw the figure. 3. In triangle XYZ, angle X is 30° and angle Z is 60°. Find X Y if YZ is 116 ft. Finding Unknowns by Ratios of Sides of Triangle 103 4. In Fig. 75, CD bisects angle C and is perpen- dicular to AB. How many degrees in angle BCD ? If CD is 100 cm., how long is DB ? In the right triangle ABC, angle B is 60°, angle A is 30°, and BC is 50 ft. Find AC In triangle XYZ, angle X is 30°. What do ZY you know about the ratio ^77^? State defi- nitely when this ratio is equal to .58. Fig. 76 Section 41. It is convenient to name important ratios. Since it is helpful to use the ratios of the various sides of a right triangle, very frequently in finding unknown dis- tances, each is given a definite name. The ratio of the "side opposite " an acute angle to the "side adjacent" is called : io4 Fundamentals of High School Mathematics THE TANGENT OF THE ANGLE Its abbreviation is tan. Thus, in the above examples the tangent of an angle of 30° is constant; it is approxi- mately .58. EXERCISE 42 1. Construct a right triangle such as Fig. 77, with AB equal to 4 cm. and angle A equal to 40°. Then measure BC and from that Q find the tangent of an angle of 40°. Compare results with those of other members of the class. 2. In a similar way find the tangent of an angle of 50°. (Use AB as 4 cm.) Also find, the tangent of each of the fol- lowing angles : 60°, 70°, and 20°. Section 42. Summary of steps in finding the tangent of an angle. These examples show how to find the tangent of any angle. TJiree steps are necessary ; namely : (1) meas- ure the side opposite the particular angle ; (2) measure the side adjacent the angle; (3) divide the first number ob- tained by the second. To do this, however, for angles of all sizes from very small to very large, would require a great deal of labor, and probably give, for a great many of you, inaccurate results. To save this trouble, and at the same time get very accurate results, these ratios or tangents have been computed very carefully and compiled in a table like Table 4. (See Table of Tangents on page 105.) Fig. 77 Finding Unknowns by Ratios of Sides of Triangle 105 Table 4 TABLE OF SINES, COSINES, AND TANGENTS Numerical Values of the Tangents, Cosines, and Sines of the Angles from 0° to 90° Inclusive Deg. tan cos sin Deg. tan cos sin .000 1.000 .000 46 1.04 .695 .719 1 .017 .999 .017 47 1.07 .682 .731 2 .035 .999 .035 48 1.11 .669 .743 3 .052 .999 .052 49 1.15 .656 .755 4 .070 .998 .070 50 1.19 .643 .766 5 .087 .996 .087 51 1.23 .629 .777 6 .105 .995 .105 52 1.28 .616 .788 7 .123 .993 .122 53 1.33 .602 .799 8 .141 .990 .139 54 1.38 .5S8 .809 9 .158 .988 .156 55 1.43 .574 .819 10 .176 .985 .174 56 1.48 .559 .829 11 .194 .982 .191 57 1.54 .545 .839 12 .213 .978 .208 58 1.60 .530 .848 13 .231 .974 .225 59 1.66 .515 .857 14 .249 .970 .242 60 1.73 .500 .866 15 .268 .966 .259 61 1.80 .485 .S75 16 .287 .961 .276 62 1.88 .469 .883 17 .306 .956 .292 63 1.96 .454 .891 18 .325 .951 .309 64 2.05 .438 .899 19 .344 .946 .326 65 2.14 .423 .906 20 .364 .940 .342 66 2.25 .407 .914 21 .384 .934 .358 67 2.36 .391 .921 22 .404 .927 .375 68 2.48 .375 .927 23 .424 .921 .391 69 2.61 .35S .934 24 .445 .914 .407 70 2.75 .342 .940 25 .466 .906 .423 71 2.90 .326 .946 26 .488 .899 .438 72 3.08 .309 .951 27 .510 .891 .454 73 3.27 .292 .956 28 .532 .883 .469 74 3.49 .276 .961 29 .554 .875 .4S5 75 3.73 .259 .966 30 .577 .866 .500 76 4.01 .242 .970 31 .601 .857 .515 77 4.33 ]225 .974 32 .625 .848 .530 78 4.70 .208 .978 33 .649 .839 .545 79 5.14 .191 .982 34 .675 .829 .559 SO 5.67 .174 .985 35 .700 .819 .574 81 6.31 .156 .988 36 .727 .809 .588 82 7.12 .139 .990 37 .754 .799 .602 83 8.14 .122 .993 38 .781 .788 .616 84 9.51 .105 .995 39 .810 .777 .629 85 11.4 .087 .996 40 .839 .766 .643 86 14.3 .070 .998 41 .869 .755 .656 87 19.1 .052 .999 42 .900 .743 .669 88 28.6 .035 .999 43 .933 .731 .682 89 57.3 .017 .999 44 .966 .719 .695 90 Inf. .000 1.000 45 1.000 .707 .707 io6 Fundamentals of High School Mathematics EXERCISE 43 FINDING ANGLES AND TANGENTS FROM THE TABLE OF TANGENTS Find, from Tabic 4, each of the following: 1. tan 4i>°. 2. The angle whose tangent is .58. 3. tan 57°. 4. The angle whose tangent is .94. 5. tan 14°. 6. The angle whose tangent is f . 7. tan 25°. 8. The angle whose tangent is f . 9. tan 45°. EXERCISE 44 EXAMPLES WHICH INVOLVE THE USE OF THE TANGENT OF AN ANGLE 1. Illustrative example. The brace wire AC of a telephone pole BC, Fig. 78, makes with the ground an angle of 62°. It enters the ground 15 ft. from the foot of the pole. Find the height of the pole BC. Solution : BC AB BC_ 15 BC= 15 x 1.88 = 28.2. = tangent 62°. = 1.88 (from the Table). 2. The angle of elevation of the top of a tree, from a point 75 ft. from its base (on level ground), & -1 is 48°. How high is the tree ? Fig. 78 Finding Unknowns by Ratios of Sides of Triangle 107 3. From a vertical height of 1500 yd. a balloonist notes that the angle of depression of the enemy trench is 51°. Find the distance from the trench to the point on the level ground directly below the balloonist. Make a drawing. 4. The angle of elevation of an aeroplane at point A on level ground is 44°. The point B on the ground directly beneath the aeroplane is 450 yd. from A. How high is the aeroplane? 5. If a flagpole 42 ft. high casts a shadow 63 ft. long, what is the angle of elevation of the sun ? 6. In Fig. 79, CD is per- pendicular to AB. Find AD if angle ^ = 60° and CD = 20. 7. From the point of observation on a mer- chant vessel, the an- gle of depression of the periscope of a submarine was 17°. How far was the submarine from the merchant vessel, if the observer was 40 ft. above the water ? 8. Turn back to page 79 and solve problem 15 by this method. How do your results compare with those obtained by scale drawings ? 10S Fundamentals of High School Mathematics II. THE COSINE OF AN ANGLE Section 43. The ratio of the "side adjacent " the given angle to the hypotenuse of the triangle, i.e. the COSINE. In the previous section we found that the ratio of the "side opposite" to the "side adjacent" an acute angle of a right triangle is always constant for any particular angle. This enabled us to find the length of the sides and the size of the acute angle. Now we come to another fact about right triangles. Let us examine a problem which cajinot be solved by the use of the tangent. In Fig. 80, BC repre- sents a telephone pole, AC an anchor wire, and ABthe dis- tance from the foot of the pole to the point at which the wire enters the ground, 20 ft. The wire makes an angle of 30° with the ground. How long is the wire? Fig. 80 Clearly, AC cannot be found by means of the ratio which we called the tangent, because the tangent of 30° makes use only of BC and AB y and we must get a ratio which contains AC, Therefore, to solve this problem we shall have to learn how to use the ratio of tJie "side adjacent" AB the 30° angle, to tJie hypotenuse, or — - . Au Finding Unknowns by Ratios of Sides of Triangle 109 EXERCISE 45 EXPERIMENTS TO DETERMINE THE NUMERICAL VALUE OF THE RATIO THE "SIDE ADJACENT" A 30° ANGLE /£ THE C0SINB THE HYPOTENUSE OF THE TRIANGLE 1. Measure the length of b x and c 1 in Fig. 81. Then compute the ratio — to two decimals. 2. Draw any other triangle similar to Fig. 81, but with much longer sides. Find, as in Example 1, the ratio of the side adjacent the 30° angle, to the hypotenuse. Compare your result with that of Example 1. Fig. 82 3. In Fig. 82, EF and GH are perpendicular to AB. Why does AH = AF = AB AG AE AC i to Fundamentals of High School Mathematics Section 44. The COSINE of a particular angle is CON- STANT. This shows that the ratio of the " side adjacent " a 30° angle to the hypotenuse of one right triangle is equal to the same ratio in any other right triangle which has an acute angle of 30°. For this reason, you would get the same numerical value for - 1 in Examples 1 and 2, if it c l were not for errors in measurement. Therefore, just as in the case of the tangent, so the " side adjacent " 30° ansrle . , cosine, i.e. '- - — , is always constant, hypotenuse when the angle is 30°. It is approximately .86. The right triangles may differ in size and position, but as long as they are similar (that is, so long as the acute angles we are dealing with are the same size), this ratio does not change. EXERCISE 46 PROBLEMS SOLVED BY APPLYING THE CONCLUSION ARRIVED AT ABOVE; NAMELY, THE RATIO OF THE "SIDE ADJACENT" A 30° ANGLE TO THE HYPOTENUSE IS .86. 1. Illustrative example. The anchor wire AC, of a telephone pole, meets the ground 20 ft. from the foot of the pole, making an angle of 30° with the ground. Find the length of the wire AC. Solution: ^=.86. (Why?) r 20 .86. ^s' .86 h, or h = 23.2 ft. y^ kS* 30° 1 _ B Fig. 83 Finding Unknowns by Ratios of Sides of Triangle 1 1 1 2. The rope, AC> of the flagpole, BC, makes an angle of 30° with the ground, at a point 42 ft. from the foot of the pole. How long is the rope ? Make a drawing. 3. The angle of elevation of the top of a tree from a point A y on level ground, 100 ft. from the base of the tree, is 30°. What is the distance from A to the top of the tree ? 4. In the right triangle ABC, AB is 64 cm. and ZA = 30°. Find AC -64 cm. Fig. 84 5. Draw a right triangle such that angle A = 60° and the hypotenuse AC =60 cm. From this could you find BC? 6. The angle of depression of a boat, from the top of a cliff, is 30°. Find the distance from the observer to the boat, if the boat is 400 ft. from the foot of the cliff. These examples have been solved by using the ratio of the "side adjacent" an acute angle of 30° to the hypotenuse, or, as we shall call it from now on, by using the cosine of the angle. The abbreviation for cosine is cos. Thus, , D ratio of " side adjacent "ZB cosZ B = —± — • hypotenuse of the triangle 112 Fundamentals of High School Mathematics EXERCISE 47 l. Construct a right triangle similar to Fig. 85, with A = 40° and AB = 4 cm. Then measure c x and compute the ratio ill. By comparing your result with cos 40° as given in the table, see if you are within .05 of the correct result. C l< — 1^=4 cm. Fig. 85 B 2. How would you construct or draw the cos of a 60° angle ? of an 80° angle ? Read from the table of cosines : (a) cos 67°. (&) The angle whose cos is .258. (c) cos 45°'. (d) The angle whose cos is .573. (e) cos 2°. (/) The angle whose cos is .707. \g) cos 89°. (//) The angle whose cos is .629. A surveyor desires to measure the distance EC across a swamp. He surveys the line BA per- pendicular to BC. He extends this line BA until he can measure from A to C. If AC is 400 ft. Finding Unknowns by Ratios of Sides of Triangle 113 7. and angle C is 55°, show how he would com- pute the length of BC. Find BC A boy observes that his kite has taken all the string, 750 ft. Assuming that the string is straight and that it makes an angle of 34° with the ground, how far on level ground is it from the boy to the point directly below the kite ? The angle of elevation of the top of a tent pole, from a point 43.2 ft. from the foot of the pole, is 32°. Find the distance from the point of ob- servation to the top of the pole. Figure 87 is a risfht triangle. angle A is 42° and AC = 61 is the cosine of angle A ? Find AB if Hint : What Fig. 8' ii4 Fundamentals of High ScJiool Mathematics 8. How long a rope will be required to reach from the top of a flagpole to a point 19 ft. from the foot of the pole (on level ground) if the rope makes an angle of 63° with the ground ? 9. The aogle of depression of a boat from the top of a cliff is 37° when the boat is 1260 ft. from the foot of the cliff. Find the distance from the boat to the top of the cliff. 10. Find angle A if AB is 27 and AC is 48. Hint : What is \\ with respect to angle A ? ll. 12. Fig. 88 A man starts at O and travels in a direction which is 48° east of a north-south line. How far due north of O will he be when he is 26 miles from O ? From the table find the cosine of 32°. Then find the cosine of an angle twice as large as 32°, and see if it is twice as large as the cosine of '■\'l . Does the cosine of an angle change or vary in the same way that the angle changes or varies ? Finding Unknowns by Ratios of Sides of Triangle 115 III. THE SINE OF AN ANGLE Section 45. The ratio of the " side opposite " an acute angle to the hypotenuse, or, the SINE of the angle. We have now used two particular ratios of the sides of a right triangle, the tangent and the cosine. By using them we were able to determine unknown lines and angles. But these two ratios are not sufficient to find any side or Fig. 89 any angle of a right triangle. For example, we have no ratio which involves BC and AC in Fig. 89. This brings us to the third (and last) important ratio : the " side opposite" an acute angle the hypotenuse In the same way as before, we can show that the numeri- cal value of this ratio is constant for any given angle. Having discussed the tangent and cosine so completely, it is unnecessary to take the trouble to construct or to com- pute the value of this ratio. The numerical values of this ratio, for all acute angles, are given in the table of SINES. n6 Fundamentals of High School Mathematics FIG. 90 EXERCISE 48 EXAMPLES SOLVED BY MEANS OF THE SINE OF AN ACUTE ANGLE 1. A man travels from O in a direction which is 50° east of a north-south line. How far is he from the north-south line when he has trav- eled 60 miles from the starting point, O ? 2. How far was the man from an east-west line through the point O? 3. To what height, on a vertical wall, will a 38-foot ladder reach, if it makes an angle of 58° with the ground ? 4. An aviator, 4200 yd. directly above his own lines, takes the angle of depression of the enemy's bat- tery. What must be the range of the enemy machine guns to endanger him, if the angle of depression is 29° ? 5. What ratio gives the sine of Z A in this figure ? co- sine C? If A is 60°, what is C? Compare sine 60° with cosine 80°, from the table. State in words your conclusion. FlG - 91 Finding Unknowns by Ratios of Sides of Triangle 117 REVIEW EXERCISE 49 In this list of problems you will have to decide for your- self whether to use the tangent, the cosine, or the sine. Make a drawing for each problem ; indicate the parts that you know, and the part you are to find. 1. A flagpole 50 ft. high casts a shadow 80 ft. long. What is the angle of elevation of the sun ? What time of year is it ? 2. A searchlight on the top of a building is 180 ft. above the street level. Through how many degrees from the horizontal must its beam of light be depressed so that it may fall directly on an object 400 ft. down the street from the base of the building ? 3. From the top of a cliff 120 ft. above the surface of the water, the angle of depression of a boat is 20°. How far is it from the top of the cliff to the boat ? 4. At a time when the sun was 55° above the hori- zon, the shadow of a certain building was found to be 98 ft. long. How high is the building ? 5. A 40-foot ladder resting against a building makes an angle of 53° with the ground. Find the dis- tance from the foot of the ladder to the building, and the distance from the top of theJadder to the base of the building. 6. A man starts at O and travels in a direction which is 24° west of a north-south line through O y at the rate of 80 miles per day. At the end of 4 days how far north is he from an east-west n8 Fundamentals of High School Mathematics line through O? How far west is he from a north-south line through O ? 7. What direction will a boy be from his starting point if he goes 40 miles due north and then 18 miles due east? 8. The gradient or slope of the railroad which runs up Pike's Peak is, in some places, 18 %, i.e. in going 100 ft. horizontally it rises 18 ft. What angle does the road make with the horizontal ? CHAPTER VIII HOW TO SHOW THE WAY IN WHICH ONE VARYING QUANTITY DEPENDS UPON ANOTHER Section 46. Quantities that change together. We have already seen that there are many illustrations of quantities that change together. The amount of money paid out for rent at $30 per month changes with, or depends upon, the number of months ; the time required to walk a certain distance, say 10 miles, changes as, or depends upon, the number of miles one walks per hour. In other words, there are varying quantities which are so related that a change in the value of one of tliem causes a change in tlie value of the otJier. This chapter will deal with quantities which change together. In addition to what you already know about these varying quantities, we shall now study just how these quantities vary. For example, does an increase in the value of one varying quantity cause a corresponding in- crease in the related quantity ? Or does an increase in one varying quantity cause a corresponding decrease in the other? Can- these be expressed (1) graphically, or (2) by tables, or (3) by formulas? These are the points which will be studied in the chapter. Section 47. Variables and constants. In our study of time, rate, and distance problems we saw that the distance traveled by a train running at any given rate changes or varies as the time which it has been running cJianges or varies. If a train runs at the rate of 40 miles per hour, its movement is described by the equation d=4:0t. In this equation, d and / change as the train progresses along its journey. The value of d depends upon the value 119 V i2o\j Fundamentals of High School Mathematics of t. This means that the distance and time are variables, while the rate is constant. Table 5 shows the tabular method of representing the relation between these related variables. This shows that Table 5 If the no. of hrs. is 1 2 3 4 5 8 10 IS 20 then the distance is 40 80 120 160 200320 400 600 800 a change in the time causes a change in the distance, or that a change in one variable causes a change in the related variable. EXERCISE 50 1. In the above table, does an incirase in the num- ber of hours always cause an increase in the dis- tance ? 2. In the same table, find the ratio of each distance to its corresponding time. How do these ratios compare ? Do the ratios change ? 3. A man buys a railroad ticket at 3 cents per mile. Show by the tabular method the relation between the cost and the number of miles traveled. Show from the table that as the dis- tance increases the cost increases, but that the ratio of the cost to the distance does not change. What equation will show the same thing the table shows ? 4. Write the equation for the cost of any number of pounds of sugar at 9 cents per pound. What are the variables in your equation ? Tabulate How One Varying Quantity Depends on Another 121 the cost for 1, 2, 5, 8, and 10 pounds. Show from the table that the ratio of the cost to the number of pounds does not change ; that is, it is constant, 5. A rectangle has a fixed base, 5 inches. Its alti- tude is subject to change. Tabulate its area if its altitude is 4, 6, 8, 10, and 12 inches. Com- pare the ratio of any two values of the area with the ratio of the two corresponding values of the altitude. If one altitude is three times another altitude, the one area is I times the other area. Write the equation for its area. 6. A bicyclist rides 10 miles per hour. Show, by three methods, the relation between the number of miles he travels and the number of hours required. In 6 hours he travels • times as far as he travels in 8 hours. I. DIRECT VARIATION, OR DIRECT PROPORTIONALITY Section 48. The problems in the previous exercise illus- trate direct variation, or direct proportionality. In each of the examples, one of the variables depended upon another variable for its value, and the ratio of any tzvo values of one variable was equal to the ratio of the two corresponding values of the other variable. When two variables are related in this way, one is said to vary as or to be directly proportional to the other. Thus, to prove that two varia- bles are directly proportional, or vary directly, we must show that The ratio of any two values of one variable is equal to the ratio of the two corresponding values of the other variable. [22 Fundamentals of High School Mathematics EXERCISE 51 1. Illustrative example. A man earns $ 6 per day. Show that the amount he earns is directly proportional to the num- ber of days he works. Solution : (1) A = 6 d. (We write the equation first, from the condi- tions of the problem.) (2) Tabulating : Table 6 If dis l 2 5 8 10 12 then A is 6 12 30 48 60 72 (3) Now select any two values of A, say 12 and 60, and the two corresponding values of d, which are 2 and 10. If the ratio of these two values of A is equal to the ratio of these two values of d, then in the equation A — 6 d we know that A is directly proportional to d, or that A varies directly as d. Does J5 = T %? Yes. Thus, A is directly proportional to d, or the amount a man earns at 1 6 per day is directly proportional to the number of days A x rfi he works. This is often written — = . A 2 d 2 Ai means some particular value of A, and A 2 means some other particular value of A; d\ and d> mean those particular values of d which correspond to the selected values of A\ and^4 2 - 2. Write the equation for the area of a rectangle whose base is 10 inches. Then show by select- ing particular values of A and k that the area is directly proportiojial to the altitude. In other words show that Hoiv One Varying Quantity Depends on Another 123 3. Write the equation for the circumference, C, of a circle whose diameter is D. Is C directly pro- portional to D ? Why ? 4. Show that the area of a square is directly pro- portional to the square of its side. 5. Write the equation for the area of a circle. Show that the area varies directly as the square of the radius. 6. Show that the interest on 11000 at 6% is di- rectly proportional to the time. 7. x varies directly as r, and when ,r = 1<), j' = -. Find the value of x when y = 7. 8. C varies directly as d % and when cf= 12, e= 4 V . What is d when c = 72 ? 9. Is your grade in mathematics directly propor- tional to the amount of time you spend in pre- paring your lessons ? 10. Is the cost of a pair of shoes directly propor- tional to the size ? II. INVERSE VARIATION Section 49. When quantities are inversely related to each other. In the previous exercise the varying quantities were so related in any particular problem that an increase in one variable caused a corresponding increase in the other variable. Some variables, however, are so related that an increase in one is accompanied by a corresponding decrease in the other. An example : An increase in the rate at which a train moves causes a decrease in the time required to travel a certain distance. If the train travels at the rate of 20 miles If the rate is 10 121 15 20 25 30 33| 40 50 then the time is 10 8 6f 5 4 ^2 °3 3 2* 2 r24 Fundamentals of High School Mathematics per hour, it will require 5 hours to cover 100 miles ; but if it increases its rate to 30 miles per hour, it will decrease the time so that only 3^ hours will be required to make the trip. Let us illustrate this fact more in detail by tabulating the relation between the rate and the time of a train which makes a trip of 100 miles. Note from the table- how a change in one variable, say the rate, is accompanied by a change in the other variable, the time. Table 7 This shows that an increase in the rate is accompanied by a decrease in the time. If we select any two values of the rate, say 20 and 50, and the corresponding values of the time, 5 and 2, we see that the ratio of the two values of the rate §{} is not equal to the ratio of the corresponding values of the time |. Clearly, |g does not equal |, or, to use the more general form, r t -1 does not equal -1. These ratios would be equal, however, if we should invert one of them, e.g. 20 50 r or - 5 ; l_ The fact that the ratio of any two values of one of the va- riables is equal to the inverted ratio of the corresponding How One Varying Quantity Depends on Another 125 values of the other variable leads us to say that one of them is inversely proportional X.o the other, or varies inversely as the other. This gives the following principle: One variable is inversely proportional to another when the ratio of any two values of one of them is equal to the IN- VERTED RATIO of the two corresponding values of the other. EXERCISE 52 1. The area of a rectangle is 200 sq. in. Show that the base varies inversely as the altitude, or b-i a that t- = — * b 2 a x 2. The number of men doing a piece of work varies inversely as the time. If 10 men can do a piece of work in 32 days, in how many days can 4 men do the same work ? 3. The variable y varies inversely as x, and when x = 12, y = 4. Find x when y = 16. 4. Write an equation to show that the altitude and base of a rectangle, whose area is fixed, are in- versely proportional. Section 50. Graphical method of representing inverse va- riation. Figure 92, on the following page, shows graphi- cally the relation between two numbers which are inversely proportional, or which vary inversely. It represents the base and altitude of a rectangle whose area is always con- stant, say 100 sq. ft. 126 Fundamentals of High School Mathematics (N I I i \ to QO \ i \ CM r i* 1 i 1 WCM go 1 US \ ' \ c0 ■•^. ^> k^._ ■«»•«■ — * — 1 >—. O 4- 8 12 16 20 24- 28 32 36 ^K) 44 48 52 BASE Fig. 92. The line shows the relationship between two numbers which vary INVERSELY ; in this case the relationship between the altitude and base of a rectangle whose area is constant, say 100 sq. ft. As the altitude INCREASES, the base DECREASES. To construct this graph, the following table was made : Table 8 If base is 2 4 5 6 8 io 12.5 20 then altitude is 50 25 20 16.6 12.5 10 8 5 Note that as the base increases, the altitude decreases. How does the ^raph show this relation ? In what way does this graph differ from those you have previously dealt with ? How One Varying Quantity Depends on Another 127 Show that the equation describes the relation between the base and altitude of any rectangle whose area is constant, say 100 sq. ft. EXERCISE 53 GRAPHICAL REPRESENTATION OF INVERSE VARIATION 1. The product of two variables, x and 7, is always 200. Tabulate 10 pairs of values of these vari- ables, and from the table construct a graph show- ing the way in which the variables are related. Measure values of x along the horizontal axis. 2. Some tourists decide to make a trip of 100 miles. Show graphically the relation between (1) the different rates at which they might travel, and (2) the time required at each rate. SUMMARY 1. Two related variables or changing quantities are directly proportional, or vary directly, when a change in one is accompanied by a corresponding change in the other. To test for direct variation, it is necessary to see whether the ratio of any two particular values of one variable is equal to the ratio of the two cor- responding values of the other variable. [28 Fundamentals of High School Mathematics 2. Two related variables or changing quantities are inversely proportional, or vary inversely, when an increase in one is accompanied by a corresponding decrease in the other. To test for inverse variation, it is necessary to see whether the ratio of any two values of one vari- able is equal to the inverted ratio of the two cor- responding values of the other variable. 3. The graph of direct variation is a straight line, while the graph of inverse variation is a curve. REVIEW EXERCISE 54 1. If 60 cu. in. of gold weighs 42 lb., how much will 35 cu. in. weigh ? 2. If a section of a steel beam 10 yd. long weighs 840 lb., how long is a piece of the same material which weighs 1250 lb. ? 3. At 40 lb. pressure per square inch, a given pipe discharges 160 gal. per minute. How many gallons per minute would be discharged at 65 lb. pressure? 4. A steam shovel can handle 900 cu. yd. of earth in 7 hr. At the same rate how many cubic yards can be handled in 5 hr. ? 5. A train traveling at the rate of 50 miles per hour covers a trip in 5 hours. How long would it take to cover the same distance if it traveled at the rate of 35 miles per hour ? 6. If 50 men can build a boat in 20 days, how long would it take -30 men to build it ? How One Varying Quantity Depends on Another 129 7. A wheel 28 in. in diameter makes 42 revolutions in going a given distance. How many revolu- tions would a 48-inch wheel make in going the same distance ? 8. The volume, v, of a gas is inversely proportional to its pressure, /. Write an equation showing this fact. 9. If the volume of a gas is 600 cubic centimeters (cc.) when the pressure is 60 grams per square centimeter, find the pressure when the volume is 150 cubic centimeters. 10. When are two changing quantities or variables directly proportional ? When do they vary inversely ? 11. How can you test for direct variation ? for in- verse variation ? Are x and y directly propor- tional in the equation x = 2y-\-5? 12. If you know that 7£_6 = 2£ + 24, then what is done to each side of the equation to get 7 b = 2b + 30 ? What is the next step in solving this equation ? Find the value of b. How do you check it ? 13. How can you get rid of fractions in the equation 1^+5 = 1^+29? Why is 24 not the most convenient multiplier? 14. In order to save d dollars in n years, how much would your savings have to average per month? CHAPTER IX THE USE OF POSITIVE AND NEGATIVE NUMBERS Section 51. We need numbers to represent opposite quali- ties, or numbers of opposite nature. The examples in the following exercises will illustrate what is meant by opposite qualities, or numbers of opposite nature. We shall take four different kinds of illustrations : (1) opposite numbers on a temperature scale, (2) opposite numbers on a distance scale, (3) opposite numbers to represent financial situations (" having " and " owing "), (4) opposite numbers on a time scale, to represent " time before " a beginning point and "time after." FIRST ILLUSTRATION: OPPOSITE NUMBERS ON A TEMPERATURE SCALE EXERCISE 55 1. The top of the mercury column of a thermometer stands at zero degrees (0°). During the next hour it rises 3°, and the next it rises 4°. What is the temperature at the end of the second hour ? 2. The top of the mercury column stands at 0°. During the next hour it falls 3°, and in the next it/alls 4°. What is the reading at the end of the second hour ? 3. If it starts at 0°, rises 3°, and then falls 4°, what is the reading ? 4. If it starts at 0°, falls 3°, and then rises 4°, what is the reading ? These examples show that we must distinguish two kinds of temperature readings, (1) those above zero and ('1) those below zero. People have agreed to call read- ings above zero " POSITIVE" and readings below zero 130 The Use of Positive and Negative Numbers 131 "NEGATIVES Thus, if the mercury starts at zero and rises 4°, it will be at positive 4°, or, more briefly, + 4°. But if it starts at zero and falls 4°, it will be at negative 4°, or — 4°. In the remainder of these examples you should describe the mercury readings as positive or nega- tive, rather than as above or below zero. 5. The temperature stands at zero. Its first change is described by the expression + 6°. Its next change is described by + 4°. What is the tem- perature at the end of the second change ? 6. If the temperature reading is 0°, and it makes the change — 5°, then — 3°, what is the final reading ? SECOND ILLUSTRATION: OPPOSITE NUMBERS ON A DISTANCE SCALE EXERCISE 56 1. An autoist starts from a certain point and goes east 10 miles, and then east 8 miles. How far and in what direction is he from the starting point ? 2. If he had first gone west 10 miles, and then west 8 miles, how far and in what direction would he have been from his starting point ? 3. If he had first gone east 10 miles and then west 8 miles, how far and in what direction would he have been from his starting point ? 4. If he had first gone west 10 miles, and then east 8 miles, how far and in what direction would he have been from his starting point ? 132 Fundamentals of High School Mathematics These examples show that we must distinguish between opposite distances, those cast of some starting point, and those west of the starting point. People have agreed to call distances cast of the starting point positive and distances west of the starting point negative. By this means a great deal of time can be saved, because a positive or negative number tells both the direction and the distance of a point on the distance scale, from some beginning point. Thus, on the distance scale, Fig. 93, point A is completely described by the number — 5. West A East 1 1 \ T 1 1 1 1 1 — I -20 -15 -10 -5 +5 +10 +15 +20 +25 Fig. 93. Points on a distance scale. This number, — 5, tells that the point A is 5 units west of, or to the left of, the starting point. 5. What would be the position on this distance scale of a man who starts at the zero point, goes east 60 units, and then west 15 units ? 6. Where would you be if you started at zero, went + 8 units, and then —8 units ? 7. A man starts at ; at the end of the first day he is at + 20, and at the end of the second day he is at — 10. What is the total distance he traveled ? What number will completely de- scribe his position at the end of the second day? 8. How far is it from + 9 to — 6 ? What direction is it ? Section 52. Thus, positive and negative numbers are used to distinguish between opposite qualities. The fore- going examples show that we need a brief, economical way The Use of Positive and Negative Numbers 133 to denote opposite qualities of numbers. This is done by positive and negative numbers, or, as we shall say from now on, by signed numbers. Thus, in referring to tem- perature readings, e.g. the " signed " number, + 10°, shows (1) how far and (2) in what direction the mercury stands from the zero point. In describing the location of a point on a distance scale, the " signed" number, — 6, tells how far and in what direction the point is from the starting or zero point ; that is, 6 units to the left of, or to the west of, the zero point. THIRD ILLUSTRATION: OPPOSITE NUMBERS TO REPRESENT FINANCIAL SITUATIONS Section 53. Positive and negative numbers, or SIGNED NUMBERS, are used also to describe financial situations. It has been agreed to consider money that you " have " as positive and money that you " owe " as negative. Thus, if you owe 40 cents (i.e. — 40 cents) and have 55 cents (+55 cents), your real financial situation is +15 cents. Why ? Or, if you owe 90 cents ( — 90 cents) and have 75 cents (+ 75 cents), your ;r#/financial situation is — 15 cents. FOURTH ILLUSTRATION: OPPOSITE NUMBERS ON A TIME SCALE Signed numbers are used also to distinguish " time be- fore " from "time after" a given time. For example, if time before Christ is negative, then time after Christ is positive. Thus, on the time scale below, since Christ's birth is regarded as zero, if a man was born 10 years be- EC- , . M^fffl-T^ . AP+ 25 -20 -75 -10 S O +5 +10 +15 +20 +25 +30 +35 Fig. 94. Points on a time scale. 134 Fundamentals of High School Mathematics fore Christ and lived 35 years, the distance between the points A and B would represent the period of his life. Why ? OTHER ILLUSTRATIONS OF THE USE OF SIGNED NUMBERS : FOR THE .PUPIL TO DEVELOP EXERCISE 57 1. Show how signed numbers are helpful in deal- ing with latitude; with longitude. Illustrate each one. 2. Show that signed numbers are a convenience in keeping scores in games in which you either make or lose a certain number of points. 3. Can you think of any other illustrations of opposite numbers ? EXERCISE 58 PRACTICE IN USING SIGNED NUMBERS 1. Your teacher's financial situation is —$250. What does this mean ? 2. A man's property is worth 15200 and his debts amount to $3300. How can positive and negative numbers be used to represent these amounts ? What number will describe his net financial situation ? 3. The mercury at 8 a.m. was at — 6°. If it was rising 3° per hour, where was it at 9 a.m. ? at 10 a.m. ? at 11.30 A.M.? 4. Show on a time scale that Caesar began to rule the Roman people 31 years B.C., and ruled for I "> years. The Use of Positive and Negative Numbers 135 5. What is the total number of miles traveled by a man who starts at zero on the distance scale if he is at +6 at the end of the first day, — 2 at the end of the second day, and at — 8 at the end of the third day ? 6. On the distance scale, where would you be if you started at — 4 and went east 6 miles ? 7. If your financial condition is + 60 cents, — 15 cents, and — 12 cents, what single number will accurately describe your net financial situation ? 8. What was your final score in a game in which you made the following single scores : + 15, - 8, - 10, + 14, and + 15 ? 9. Represent on a distance scale (horizontal) the point where a man would be at the end of the third day if he started at zero and walked + 6 miles on Monday, — 10 on Tuesday, and — 3 on Wednesday. 10. Find the net financial situation of a man who is worth the following: (a) + $5+f8 + $10 -$6; (b) + 6d-10d-8d+lod. Section 54. Absolute value of positive and negative num- bers. The numerical value of a positive or negative num- ber, without regard to its sign, is its absolute value. For example, the absolute value of + 6 is 6 ; of + 17 is IT ; of — 9 is 9, etc. Section 55. We need to be able to add, subtract, multiply, or divide signed numbers. Now that we see clearly the practical ways in which positive and negative numbers are used we need to be able to solve problems which contain 136 Fundamentals of High School Mathematics either kind. In all the examples which we have worked previously, only positive numbers have been used. Next, therefore, we must learn (1) how to combine signed num- bers {i.e. add them) ; (2) how to multiply them ; (3) how to find the difference between two signed numbers; and (4) how to divide signed numbers. We will take them up in that order. I. HOW TO COMBINE SIGNED NUMBERS — FINDING ALGEBRAIC SUMS Section 56. When the numbers are arranged vertically. In the example : " Find the net financial situation of a man who is worth the following: +|5, +$8, -$10, -$6," we found one signed number which described the man's net financial situation; namely, — $3. That is, we found one signed number which was the result of putting several signed numbers together. This process is called combining signed numbers, or finding the algebraic sum. Thus, to combine +4, — 2, — 6, and + 3, we must find one signed number which is the result of putting all of these together. Evidently, this must be — 1. Similarly, combining, or finding the algebraic sum of + 5 d and — 11 d f we get -6d. In each of the following examples, find the algebraic sum, i.e. find one number which will describe the result of putting all the separate numbers together. l. EXERCISE 59 + 16 2. - Id 3. + 4x 4. 5 a - $3 + 8d -Zx 4a + $4 -\d - X — 6a The Use of Positive and Negative Numbers 137 5. 6jj/ 6. + 8 7. - 10 8. + 3£ -by -5 + 13 -5b - y -6 + 7 - 8 + 5 -6b + 2b 9. 4a 10. -Zx 11. Wad 12. — 7 xy — 6a — 5x + Sx 12 ab + 11 xy - 2xy 13. + 21* 14. - 4 15. X 16. 3# -7 x + 3 2x --la + 2 x + 11 - 8 + 1 -6x la - 9a Section 57. When the numbers are arranged horizon- tally. The numbers to be combined are almost always written in a horizontal line, rather than in a vertical column. Combining these terms is done in the same way as if they were written in a vertical column. For example, 17. 6 - 8 - 5 + 11 = ? 18. -8-9 + 11 + 6 19. - Qd+ 5d+9d- 2d 20. 5 abc + 6 abc — 7 abc 21. 4/- 3/+ 6/- 9/ 22. + 8 t - 9 / - 6 1 + 12 1 23. How have you found the algebraic sum of these numbers ? 138 Fundamentals of High School Mathematics Section 58. Terms are either LIKE or UNLIKE : How to distinguish TERMS. Any algebraic expression, such as ax + b or x 1 + *&xy + 5, is made up of one or more num- bers separated from each other by + signs or by — signs. These numbers thus separated from each other are called terms. Thus, in ax + b there are two terms ; ax is one, b is the other ; while in x 2 + 2 xy — 5 there are three terms, x 2 , 2xy f and 5. Note carefully that a "term" includes everything between + or — - signs. In many algebraic expressions these terms are all like terms, and, as we learned in the previous section, can be combined or put together into one number or term which we called their algebraic sum. Thus, 4 d, + 5 d, — 8d, + 3 d are similar or like terms, and their algebraic sum is + 4 d. It is important to understand that because each letter in the expression represents the same thing these are like terms. In many cases, however, the terms of an algebraic expression are not all like terms. For example, consider : 4 boys, + 5 girls, + 8 boys, — 2 girls, or, using the initial letters of the words, 4 b + bg + 8 b — 2g. Evidently, these are unlike terms and cannot be combined into one number. However, the like terms in the expression can be combined ; that is, the 4- 4 b and +8^, giving 12 b, and the + bg and — 2g, giving + Sg. Thus the expression 4 b + bg + 8 b — 2g can be simplified or expressed more briefly by combining like terms, giving 12 b 4- 3^-. From this illustration we see that the like terms of any algebraic expression can be combined, giving a simpler, briefer expression than the original one. The Use of Positive and Negative Numbers 139 EXERCISE 60 FURTHER PRACTICE IN COMBINING SIGNED NUMBERS: NUMBERS HAVING LIKE OR UNLIKE TERMS Write in the simplest or briefest form each of the following expressions : 1. 2a + 3a — 6a + ±a 2. 5 ft. + 6 in. - 2 ft. - 4 in. 3. 7 yr. + 3 mo. — 2 yr. — 1 mo. 4. ±b + 5c - 8b ~2c + b 5. 6 a 2 + 3 a 2 - 7 a 2 + 4 a 2 6. - 2x*- 5x*- 8x* + 2*3 7. ax + 5 + 4 ax + 3 8. 3 xy + 5 ab — 7 ,17 — 11 <z£ 9. 2 £ 3 - 7 £ 3 + 5 js - #> 10. 2r+8r + 3r- 10 r 11. 4 * 2j + 5 ^ _ 8 ^ _ 3 a * b 12. _6 + 4-8+6-9 + 2 13. 5^+3-8^-4 + 4^ + 1 14. a 2 b + 4 *2j _ 6 a 2 £ 15. .27 + 3 - 8 *y - 9 + 2 4:7 + 7 16. / + 2^-8/ + 4^ + 6/ + 5^ SUMMARY OF IMPORTANT PRINCIPLES CONCERNING THE COMBINING OF SIGNED NUMBERS Section 59. You have now worked many examples in finding algebraic sums. From your experience with such examples, complete these three sentences which tell how to combine signed numbers : 140 Fundamentals of High School Mathematics 1. To find the algebraic sum of two positive numbers, l the absolute values of the numbers, and give to the result a I sign. 2. To find the algebraic sum of two negative numbers, ! the absolute values of the numbers, and give to the result a 1 sign. 3. To find the algebraic sum of two numbers having unlike signs, find the I of their absolute values, and give to the result the sign of the number having the I absolute value. II. HOW TO MULTIPLY SIGNED NUMBERS Section 60. The four ways to multiply signed numbers In arithmetic it was found that multiplication shortened the work of addition. For example, in adding 3 + 3 + 3 + 3 + 3 + 3 + 3, the result is found most easily by multiply- ing 3 by 7, because 3 is taken 7 times. So, in algebra, it is equally desirable to multiply one signed number by another. There are four different ways in which we may have to multiply signed numbers. These are : (1) plus times plus, as in the example +4 times + 2 == ? (2) plus times minus, as in the example + 4 times — 2 = ? (3) minus times plus, as in the example —4 times + 2 = ? (4) minus times minus, as in the example — 4 times — 2 = ? By considering the following problems we can tell what meaning must be given to the multiplication of signed numbers. The Use of Positive and Negative Numbers 141 A. ILLUSTRATIVE QUESTIONS BASED UPON THE SAVING AND WASTING OF MONEY EXERCISE 61 1. If you save $5 a month ( + $5), how much better off will you be 6 months from now (+ 6) ? Evidently you will be $30 better off ( + $30). Thus, + 5 times + 6 = + 30. 2. If you have been saving $5 a month ( + $5), how much better off were you 6 months ago (— 6)? Evidently you were $30 worse off (- $30) than you are now. Thus, + 5 times — 6 = — 30. 3. If you are wasting $5 a month (— $5), how much better off will you be in 6 months from now (+6)? Evidently you will be $30 worse off ( — $30). Thus, — 5 times + 6 = — 30. 4. If you have been wasting $5 a month ( — $5), how much better off were you 6 months ago (-6)? Evidently you were $30 better off (+$30). Thus, — 5 times — 6 = + 30. Summarizing : These problems based upon saving and wasting money have led to the following illustrative state- ments : 1. + 5 times + 6 = + 30! 2. + 5 times - 6 = - 30. 3.-5 times + 6 = - 30. 4.-5 times - 6 = + 30. 142 Fundamentals of High School Mathematics B. ILLUSTRATIVE QUESTIONS BASED UPON THERMOMETER READINGS EXERCISE 62 1. If the mercury is now at zero and is rising 2° per hour (+ 2), where will it be 4 hours from now (+4)? Evidently it will be 8° above zero (+ 8). Thus, + 2 times + 4 = + 8. 2. If the mercury Jias been rising 2° per hour ( + 2°) and is now at zero, where was it 4 hours ago (-4)? Evidently it was 8° below zero ( — 8°). Thus, + 2 times — 4 = - 8. 3. If the mercury is now at zero and is falling 2° per hour ( — 2°), where will it be 4 hours from now (+4)? Evidently it will be 8° below zero (- 8°). Thus, - 2 times + 4 = - 8. 4. If the mercury is now at zero and has beenfall- i?ig 2° per hour (— 2°), where zvas it 4 hours ago (-4)? Evidently it was 8° above zero ( + 8°). Thus, - 2 times —4 = 4-8. Summarizing : these problems based upon the ther- mometer have led to the following illustrative statements : 1. + 2 times + 4 = + 8. 2. + 2 times — 4 = — 8. 3.-2 times + 4 = - 8. 4.-2 times — 4 = + 8. The Use of Positive and Negative Numbers 143 A careful study of these illustrations will enable you to complete the following statements concerning multiplica- tion of signed numbers : 1. A positive number multiplied by a positive number gives as a product a ? number. 2. A positive number multiplied by a negative number gives as a product a ? number. 3. A negative number multiplied by a positive number gives as a product a ? number. 4. A negative number multiplied by a negative number gives as a product a 1_ number. 5. The product of two numbers having like signs is ? 6. The product of two numbers having unlike signs is ? II a. PARENTHESES ARE USED TO INDICATE MULTIPLICATION Section 61. Multiplication of two or more numbers is often indicated by placing the numbers within parentheses. Thus, " + 4 times -35" is often written " ( + 4)(- 35)." It is important to note that no sign or symbol is placed between the parentheses when multiplication is indicated. EXERCISE 63 PRACTICE IN MULTIPLYING SIGNED NUMBERS 1. ( + 3)(+5) 8. < r +5)(+4)(-2) 2. (+6)'(-2) 9. ( _3)(-6)(+2) 3. (+10)(-2|) 10. ( ;-4)(-10)(-3; 4. ( + 6)(-9) 11. ( :+§)(- 1) 5. (-2)(-5) 12. ( ;-fx+i) 6. (+8)(-|) 13. ( ;-7)(-6)(+2) 7. (+12)(+6) 14. ( ;-2)(-2)(-2) 144 Fundamentals of High School Mathematics 15. (- 3)( -3)(- -3) 16. (+l)( + l)( + l)+(+l) 17. ( -2)(-2) 31. (- §)(+!) 18. ( -■2)(-2)(-i 0(- •2) 32. (-■W-tt) 19. ( 3)(4)(5)(2) 33. (_2)(-2)( + 2) 20. ( + 2)( + #5) 34. ( + *)(-tt) 21. ( + 3)(+7Z>) 35. (-JX- io) 22. ( ; + (!)( + 5 ft.) 36. -5-8 23. ( -8)( + 6j) 37. -7 -21 24. ( + f)(18«) 38. (-6X-*) 25. (-12X+J) 39. 2 -2 -(-3) 26. ( IX- 18) 40. (_1)(-1)(-1)(_1) 27. ( .27)(-!) 41. (2)(-3)( + 4)(-5) 28. ( : + !)(- 25) 42. (6)(|)(-1) 29. ( :-32)(-|) 43. (10)(5)(-^) 30. ( :-iK+2±j) 44. (IX- 1) II J. HOW TO USE EXPONENTS IN MULTIPLICATION Section 62. Suppose we had to find the product of 3x 2 and 5.r 4 . It is important to keep in mind the meaning of exponents. 3 x 2 means 3 - x • x and 5 x 4 means 5 • x • x • x • x. Hence, Zx 2 times 5.r 4 or (Sx 2 )(5x A ) means 3 • x • x • 5 • x • x • x • .r, or 15 ;tr 6 . By the same reasoning, the product of + 6 x 6 and - 7 x 4 is - 42 ^ 9 . The Use of Positive and Negative Numbers 145 EXERCISE 64 PRACTICE IN USING EXPONENTS IN MULTIPLYING SIGNED NUMBERS (ORAL) 1. (5a 2 )(6a 4 ) 13. - 2y ■ 3 y 2 2. (+7£)(-9^) 14. (_5J)(-2J») 3. (+8)(2y>) 15. (-6^ 2 )(-7^) 4. (6^)(2« 2 ) 16. (—8 *»)(£*) 5. ( + Sabc)(oab) 17. ( + 10ytr)( - 2y&) 6. 4 „r 3 . 5 x 1 - 2 ;r 5 is. 16 ;z 2 £ - {\ ab») 7. j 4 • 5/ 3 19. a - b - b - a - a 8. J^. 10. r 7 20. .r 4 ■ 2* 9. 2aP.§ab 21. j • 5j/ 3 10. a 2 b-ab 2 -aW 22. ^.fe 11. ;tr • 2;r 3 23. 5 . x 1 12. - 6 a • 3 a 24. 10 j 2 - T \y . y* III. HOW TO FIND THE DIFFERENCE BETWEEN SIGNED NUMBERS: SUBTRACTION Section 63. How " differences " are found in practical work. Clerks in stores have a method of making change or of finding the difference between two numbers that is very helpful in finding the difference between two signed numbers. For example, if a customer gives the clerk 50 cents in payment for a 27-cent purchase, the clerk begins at 27 and counts out enough money to make 50 cents. If we use the same terms as were used in arithmetic, — namely, the subtrahend, minuend, and difference, — then we say, " The clerk begins at the subtrahend, 27 cents, and counts to the*minuend, 50 cents." 1 46 Fundamentals of High School Mathematics First illustrative example. To illustrate this method of finding the difference between two signed numbers, let us consider this problem : On a certain day the mercury stands at — 4° in Chicago and at -f 13" in St. Louis. How much warmer is it in St. Louis, or what is the difference between -f 13° and - 4° ? Naturally, we do the same thing the clerk does, begin at the subtrahend and count to the minuend, i.e. we count from - 4° to + 13°, giving us + 17°. The difference is called positive because we counted upward. If we counted downward, the dif- ference would be called negative. This example is written as fol- lows : + 13° minuend — 4° subtrahend + 17° difference -4- +13 Chicago St. Louis Fig. 95 a Second illustrative example. Subtract -I- 10 from — 5 by referring to the number scale. This means to find the distance from the sub- trahend to the minuend or from +10 to — 5. The distance from 10 above to 5 below is clearly 15 ; and since the direction is downward, the difference is — 15. This example is written : — 5 minuend -f 10 subtrahend — 15 difference i The Use of Positive and Negative Numbers 147 These illustrations are given merely to show that the difference between two signed numbers can always be found by counting on a number scale from the subtrahend to the minuend. The difference will be positive or nega- tive, depending upon whether the direction of counting is upward or downward. EXERCISE 65 PRACTICE IN FINDING THE DIFFERENCE BETWEEN TWO SIGNED NUMBERS: SUBTRACTION 1. +6 -4 +8 +3 + 5 -7 + 13 + 9 - 2 + 5 + 2 +10 -8 + 1 + 4 +14 2. + 14 -6° + 7 d + 10 ft. -4 in. + Sx -18 + 9 5 -2d + ±b -he - 6 ft. 10 jr 2 — 7 in. -10* 3. - Za — -3 x^y + 10 adc -11a -2b + 6e -2.T 2 + 5x s y 3x — \abc 4. -2fi + 5xf 5a -2 be + 2,fi + 11 xf -2a X be 5. From — 7 a take + 5 a. 7. From x take 1 x. 6. Take - 13 b 2 from + 2 b 2 . 8. Take - be from 2 be. IV. HOW TO DIVIDE SIGNED NUMBERS Section 64. Division is the opposite of multiplication. You will have little or no difficulty in the division of signed numbers if you understand that division is just the opposite of multiplication. For example, if 4 x 2 = 8, then |=4. In this case 8 is the dividend, 2 is the divisor, and 4 is the quotient. In signed numbers, as well as in 148 Fundamentals of High School Mathematics arithmetic, the dividend equals the quotient times the divisor. + 8-*--2 = -4;or -"±-| = -4, because (- 2)(- 4) = + 8. - 8 -f- - 2 = + 4 ; or -^| = + 4, because ( - 2)(+ 4)= - 8. + 8- + 2 = +4;or:-^ = + 4, because ( + 2)( + 4)= + 8. _8-s- + 2 = -4; or r-| = -4, because (+ 2)(-4)= -8. 4- 2 EXERCISE 66 PRACTICE IN FINDING THE QUOTIENTS OF SIGNED NUMBERS Find the quotient in each of the following : 4-10. +18 . -16 . -30 . -14 . +6 . -8 . +16 , ' +2' -2' +4' -10' +7' +2' -1' -2' + 15rf —18 ft. +16 mo.. + 25.r -10<z 2. 3' +3' -8 ' -5 ' +2 3. (21-)+(-8); (_36)-*-(-9); (- 54)^+27); (-96)-K+12); (_21)4-<-7); (-60)^( + 12). , 10.r3 2iy 18 tf> 12 r 4 16/ 3 u 4 - IT 5 3/ ; 6* 5 ^ ; ifp' Howcan y° u prove each of these ? + 20/ . -27**. -34£¥= , -50^y . -72/z^ ' -4j ' -9x*' -17£ 2 ' + 25jj/' -6j/w 5 ' The Use of Positive mid Negative Numbers 149 EXERCISE 67 COMPLETING STATEMENTS ABOUT DIVISION 1. A positive number divided by a positive number gives as a quotient a 1 number. 2. A negative number divided by a positive number gives as a quotient a 1 number. 3. The quotient of two numbers having like signs is L_. 4. The quotient of two numbers having unlike signs is • EXERCISE 68 A REVIEW OF ADDITION, MULTIPLICATION, SUBTRACTION, AND DIVISION OF SIGNED NUMBERS This is a very difficult but a very important exercise. 1. From the sum of 2 a and — 5 a take the differ- ence between — 3 a and + 8 a. 2. Add the product of — 3 and + 5 to the quotient of - 18 and - 2. 3. Take the sum of — 7 b and + 4 b from the dif- ference between — 6 b and + 11 b. 4. To the quotient of — 21 and + 3 add the product of — 6 and + 7. 5. From the sum of 7 / and — 10 / take the differ- ence between — 4 i and + 11 A 6. Add the product of — 6 and + 9 to the quotient of - 28 and - 4. 7. Take the sum of — 9 x and + Sx from the dif- ference between — 5x and + 13;tr. 150 Fundamentals of High School Mathematics 8. To the product of — 7 and +11 add the quo- tient of — 33 and + 3. 9. From the sum of 8<r and — 14 <; take the differ- ence between — be and + 16*:. 10. Add the product of — 12 and + 5 to the quo- tient of - 32 and - 4. 11. Take the sum of — 8j and 3j/ from the differ- ence between — 7 y and + 12j/. 12. To the product of — 8 and + 9 add the quotient of - 36 and + 6. 13. From the sum of 12 b and — 16 b take the dif- ference between —lb and + 8 b. 14. Add the product of — 8 and + 9 to the quotient of — 40 and — 5. 15. Take the sum of — 11 1 and 7 / from the differ- ence between — 9 / and + 10 t. 16. To the product of — 6 and + 13 add the quo- tient of - 42 and + 7. SUMMARY This chapter shows the need of signed numbers. It teaches how to : 1. Combine or add signed numbers. 2. Find the difference between two signed numbers; that is, to subtract one signed number from another. 3. Find the product of signed numbers. 4. Divide one signed number by another. The Use of Positive and Negative Numbers 151 REVIEW EXERCISE 69 1. The formula h = 25 + 1 (G — 4) is used to deter- mine the proper height of the chalk trough in a schoolroom. If h stands for the height in inches, and g stands for the number of the grade, find the height for Grade VIII; that is, when g= 8. What is the proper height for a third-grade room ? 2. Evaluate the expression ab 2 + a 2 b if a = 2 and b = -3. 3. Show that the sum of any two numbers having unlike signs, but the same absolute value, is zero. Give some illustrations. 4. In a class of 25 pupils, 2 were conditioned and 6 failed. Express the ratio of the number of pupils that succeeded to the total humber in the class. What percentage is this ? 5. The ratio of y + 1 to 9 is equal to the ratio of y + 5 to 15. Find y. 6. The number of posts required for a fence is 84 when they are placed 18 feet apart. How many would be needed if they were placed 12 feet apart ? 7. If I am now x years old, what does the follow- ing expression tell about my age : 2 x + 5 = 55 ? CHAPTER X THE COMPLETE SOLUTION OF THE SIMPLE EQUATION Section 65. What we have already learned about the equation. Since the equation is the most important opera- tion in mathematics, we must be able to solve quickly and accurately equations of any kind. Thus far we have learned two very important facts about equations : 1. That if we do anything to one side of an equation, we must do the same thing to the other side. 2. That an equation is SOLVED when a value of the un- known is found which satisfies the equation ; that is, one which makes the numerical value of one side equal to the numerical value of the other side. Furthermore, we have learned: (1) how to solve simple equations of the type 6 b + 3 = 45, or, c + 5 c = 20 + c , etc. ; (2) how to get rid of fractions in an equation, e.g. of the type l^r+l^r — 1 = 3; (3) how to solve word problems, first by translating them into equations and second by solving the equations. These methods, which you have now mastered, are important first steps in the more important problem of learning how to solve equations of any kind. That is your task in studying this chapter. I. SOLVING KQUATIONS WHICH CONTAIN NEGATIVE NUMBERS Section 66. There are just two more steps that we must learn in using equations. First, we must be able to solve equations which contain negative numbers ; second, we must be able to solve equations which contain parentheses. 152 Complete Solution of the Simple Equation 153 Negative numbers occur very commonly in equations. The following examples illustrate this fact. EXERCISE 70 Write as equations, and solve each of the following examples : 1. What number multiplied by 7 equals — 28? 2. What number multiplied by —5 equals 20? 3. If a certain number be added to 13, the result is 8. Find the number. 4. A certain number increased by 10 equals — 5. Find the number. 5. If 7 be subtracted from a certain number, the result is — 3. What is the number ? 6. If negative four times a certain number gives 22, what is the number ? These examples show how negative numbers occur in equations. Throughout the remainder of the work, equa- tions which are satisfied by negative numbers will occur very commonly. The next exercise contains many exam- ples of this kind. EXERCISE 71 SOLUTION OF EASY EQUATIONS WHICH CONTAIN NEGATIVE NUMBERS Solve each of the following equations. You should be able to tell exactly what must be done to each side of the equation. 1. ;r+5 = 3 5. 2 x+lQ = 2 2. 2j=-16 6. -4^ = 12 3. £ + 7 = 2 7. 10j>/ + 2 = -18 4.-3 = 15 8. 2 £-1 = 9 154 Fundamentals of High School Mathematics 9. Throe limes a certain number, increased by 10, gives l>. What is the number ? 10. If twice a certain number is added to 16, the result equals the number increased by 6. Find the number. IX. V+ •>="!->•+ Vr+2. o 4 b 12. 12 -2*= 8. 13. The sum of two thirds of a certain number and three fourths of the same number is — 17. Find the number. A new kind of equation. 14. -2*- 12 = 5* -40. The equations which you have just solved are of the kind in which you can easily see what to do to each side. With examples like 14, however, in which both knozvns and unknowns occur on each side and which include nega- tive numbers on one or both sides, we need special and systematic practice. Section 67. We need to get "knowns" on one side and " unknowns " on the other. Just as clerks in stores always place the known weights on one scale pan and the un- known weights on the other scale pan, so we, in solving equations, always get the known numbers, or terms, on one side of the equation, and the unknown terms on the other side. Usually we get all the unknown terms on the left side, and all the known terms on the right side. Thus, in the equation above, -2* -12 = 5* -40, we do not want — 12 on the left side. Therefore we get rid of the known on the left side by adding + 12 to each Complete Solution of the Simple Equation 155 side, giving the equation — 2.;r = 5;r — 28. We also do not want the 5x on the right side. Therefore we subtract 5x from each side, giving the equation — lx = — 28. Dividing each side of this equation by — 7, we find that * = 4. Section 68. Equations should be solved in a systematic order. In learning to solve equations which require several steps, pupils make many mistakes because their work is not arranged in a set order. For some time to come, therefore, you will find it very important to use a form like the fol- lowing in solving equations : Illustrative example. Solve the equation - 2 x - 12 = 5 x - 40. (1) Adding + 12 to each side, gives - 2 x = 5 x - 28. (2) Subtracting 5 x from each side, gives - 7 x = - 28. (3) Dividing each side by — 7, gives x = 4. (4) Check : Substituting 4 for x to check the result, gives - 8 - 12 = 20 - 40. _ 20 =-20. EXERCISE 72 Solve each of the following examples, writing out each step exactly as in the solution of the illustrative example : 1. -3*- 8 = 8*- 30 8. 14 = 2j + 20 2. 5j/-6 = 9j/ + -i2 9. _7tf + 4=+8tf-41 3. -6£ + H = 2# + 43 10. _2.r-T=-8.r-19 4. jr- 20 = 50 -ft* 11. +5>-3 = 8^-16 5. _2r + 10 = 4 12. 5+2jk=0 6. 10 -8x = -20 13. 10.r + 22 = 12 7. 6-4j/=2 14. = 4.r + 20 156 Fundamentals of High School Mathematics II. HOW TO SOLVE KQUATIONS WHICH CONTAIN PARENTHESES Section 69. We saw in the last chapter that parentheses were used to indicate multiplication. Thus, to show that — 4 is to be multiplied by — 6, we use the parentheses, as follows: ( — 4)( — (J). Multiplication is usually indicated in this way. Take this example to illustrate the way in which parentheses will be used in a practical way : A. Illustrative example. Oranges cost 10 cents more per dozen than lemons ; the cost of four dozen lemons and two dozen oranges is $ 2. What is the price per dozen of each? Z?. Solution : Let x = no. cents per dozen paid for lemons ; then x + 10 = no. cents per dozen paid for oranges; then 4x = no. cents paid for 4 doz. lemons; and 2(x + 10) = no. cents paid for 2 doz. lemons. Therefore 4 x + 2(x + 10) = no. cents paid for both; or 4jt + 2(x+ 10)= 200. Note the use that is made of parentheses; that is, to show that the expression x + 10 must be multiplied by + 2. Per- forming this multiplication, (1) or removing parentheses, gives 4x + 2x + 20 = 200. (2) Combining terms gives 6at+ 20 = 200. (3) Subtracting 20 from each side gives 6x = 180. (4) Dividing each side by 6 gives x = 30 cents per dozen for lemons, and therefore x + 10 = 40 cents per dozen for oranges. (5) Substituting in the original word-statement 30 cents and 40 cents respectively for the cost of lemons and oranges enables us to check the result : 4.*. 30+ 2>5.40 = >52 Complete Solution of the Simple Equation 157 EXERCISE 73 PRACTICE IN SOLVING EQUATIONS WHICH CONTAIN PARENTHESES Solve and check each of the following equations : 1. 2(.r + 10)=42 5. 2(*-3) + 3(;r-2) = 8 2. 5(j - 2)= 15 6. 5 b -f 2(4 - 6)= 32 3. 3(2 £- 4) = 18 7. -3.r + 6(.r-4)=9 4. 4^ + 5(.r+2) = 46 a -7£ + 4(2£-3) = 16 Note that in all the foregoing examples the number be- fore the parenthesis has been positive. If negative num- bers occur, however, we proceed just the same, remember- ing how to multiply a negative number. Illustrative example. Solution of an equation involving RE- MOVAL OF PARENTHESES. 9. Sx— 2(2x- 7) = x + 8. The expression 2 x — 7 is to be multiplied by — 2. (1) Performing this multiplication, or removing parentheses, gives 8jc-4jc + i4 = x + 8. (Why is it +14?) (2) Combining terms gives 4x + 14 = x + 8. (3) Subtracting -f x from each side gives 3x4- 14 = 8. (4) Subtracting + 14 from each side gives 3x=-6. (5) Dividing each side by 3 gives x=-2. (6) Check : Substituting — 2 for x throughout the equation gives -16-2(-4-7)=-2 + 8. - 16 + 8 + 14 = - 2 + 8. 6 = 6. 1 58 Fundamentals of High School Mathematics EXERCISE 73 (continued) 10. 5£-3(4-2*)=r2£ + 42 11. (i(.r-:i)-4(x + 2)=4-.r 12. 7(6 - 2)- 2(3 + d)= 13. 4(2^ -5)+ 15 = 3^ + 10) 14. <)j-8(2j/-4)=6 Section 70. A difficult form of multiplication. A form of multiplication (as shown by parentheses) that gives pupils difficulty is the kind represented by — (4 — 5;r) in the equation : 15. 5-2(*-6)=-(4-5;r) When no multiplier appears immediately before the parentheses, the multiplier 1 is understood. Therefore in this case the multiplier is — 1. It is just as though the right side of the equation read -1(4-5^). Illustrative example. Therefore the complete set of steps required to solve this equation includes: (1) Removing parentheses gives 5 -2x + 12=-4 + 5x. (Whyisit + 5x?) (2) Combining terms gives -2x + 17=-4 + 5x. (3) Subtracting + 17 from each side gives -2x=-21+5x. (4; Subtracting — 5 x from each side gives -7jc = - 21. (5) Dividing each side by — 7 gives x = 3. (6) Substituting 3 for x, throughout the original equation to check the result, gives 5-2(3 - 8) = -(4- 53). or, 5 -6 + 12 =-4 + 15. 11 = 11. Complete Solution of the Simple Equation 159 There are two important and difficult points in this last example. First, you should note that in the expression 5 — 2(x — 6) the — 2 is not to be subtracted from the 5. The expression in parentheses must be multiplied 'by — 2. Second, if no multiplier is written before the parentheses, as in the expression — {\ — bx\ it is understood that the multiplier is 1. In this case it is — 1. If there had been no sign before the parentheses, as (1 — 5x), the multiplier would be understood to be + 1. EXERCISE 73 {continued) 16. lx-{x — 4) =,25 21. 2 £-7(3 -£)=£+ 8 17. _5 J/ _(2-j)=18 22. 1(2* + 3)= -17 18. 6x-(x+l)= -2,r + 35 23. —1(6 — 2*)= 43 19. 5 -2(^-4)= 23 24. -(6-2jt)=24 20.' 7 - 12(3 - 0)= 31 25. 16 = (2 x + 1) A. SUMMARY OF THE STEPS REQUIRED TO SOLVE EQUATIONS WHICH CONTAIN PARENTHESES Section 71. Look back to the illustrative examples, 9 and 15, and compare the steps in the solution that is worked out for each one with the steps in the solution in eacl) of those which you have just worked. You will note that to solve such an equation the following steps are always included : I. Removing the parentheses {i.e. multiplying). II. Combining like terms on each side. III. Getting rid of all known terms on one side and all unknown terms on the other side. IV. Dividing each side by the coefficient of the unknown, to give the numerical value of the unknown. V. Substituting the obtained value in the original equation to check the result. 160 Fundamentals of High School Mathematics III. HOW TO SOLVE EQUATIONS WHICH CONTAIN FRACTIONS Section 72. If, however, we should be given the equa- tion 2(x - 3) _ x - (> = x _ 16 5 4 2 5' we need to add another step ; namely, Getting rid of fractions by multiplying each term by the most convenient multiplier. Illustrative example. 1. Solution of an equation which involves getting rid of fractions. 2(s-3) x-6 = x 16 5 4 2 5* (1) Removing parentheses gives 2x - 6 s - 6 _s 16 5 4 2 5 * (2) Multiplying each term by the most convenient multiplier, 20, gives 20— — -20— =20 5 -20 r (3) Reducing fractions gives 8x-24-5jt + 30 = 10 x- 64. (Why +30?) (4) Collecting terms gives 3x -j-6 = 10jc-64. (5) Subtracting 6 from each side gives 3x = 10 jc — 70. (6) Subtracting 10 x from each side gives -7x = -70. (7) Dividing each side by — 7 gives x = 10. (8) Substituting the obtained value of the unknown to check the result. Complete Solution of the Simple Equation 161 B. SUMMARY OF THE STEPS REQUIRED TO SOLVE EQUATIONS WHICH CONTAIN BOTH PARENTHESES AND FRACTIONS The solution of this example illustrates all the steps that are ever included in solving simple equations. The com- plete list now includes : I. Removing parentheses. II. Getting rid of fractions by multiplying "each* term by the most convenient multiplier. III. Combining like terms. IV. Getting rid of all the known terms on one side and all the unknown terms on the other side. V. Dividing each side by the coefficient of the un- known. VI. Substituting the obtained value in the original equation to check the result. All these steps are not required unless the equation includes parentheses and fractions. Now that we have learned all of the steps that are neces- sary in solving simple equations we need to practice so as to be very proficient in this work. Nothing is more im- portant in high school mathematics. The next exercise is included to provide that practice. EXERCISE 74 PRACTICE IN THE COMPLETE SOLUTION OF EQUATIONS 3(.r-4) = .r + 8 x ,r-4 _ll 7 3 '25 10 What step in the above list is omitted m working this example ? io 2 Fundamentals of High School Mathematics 3. 4. f>(.r-2)_2(.r + 1) = 7 3 •") 2(3.r + 1) (.r+2) = 13 7 14 14 3^ 2(6-10) 5J-19 4 : 5 3 6. .](4.t- + 8)-§(6,r-9)=0 7. 1(6* -9)- f(8;r+ 4)= -13 8. 5(,-2) 7(2, - 3) _ 12 3 4 ^ T 4 g 2(* + 8) 3(.r-8) -3(2*+ 8) 7 2 1 10. 3(4-,)= 5(6 -3,) 11. x x x — 2 .r 2 3 6 36 12. 2£ 3£ b-b £-37 5 4 10 20 TTTi: ALGEBRAIC SOLUTION OF WORD PROBLEMS Section 73. Review of important steps in translating word problems into algebraic statements. We have taken a great deal of time to learn how to solve any kind of simple equation because we need to be able to use equa- tions skillfully in solving actual problems later. The problems as a rule will not be stated for us, in algebraic or equational form, all ready for solution. They will be stated merely in words. First, therefore, we shall always have to translate the word problem into an equation. Be- yond this first step the work is the mere solution of the equation. Complete Solution of the Simple Equation 163 Our second principal task in this chapter, therefore, is to become skillful in translating word problems into algebraic form. We learned in our work with Chapter III the important steps in translating word problems. Since we are to learn in the next few lessons how to translate a great many different kinds of word statements, let us re- view these steps here : First step:. See clearly which things in the prob- lem are known and which are un- known. Second step : Represent one of the unknowns, most conveniently the smallest one, by some letter. Third step : Represent all of the others by using the same letter. Fourth step : By careful study of the relations between the parts of the problem, ex- press the word statement in algebraic form. Sometimes this will mean an equation and sometimes not. For the next few lessons, therefore, you will work many word problems. The exercises are included to give prac- tice in translating many different kinds, so that you will be able to use the method in solving any kind that you may happen to meet later. For convenience they will be arranged by types, examples of the same type being studied together. I. PROBLEMS IN WHICH A NUMBER IS DIVIDED INTO TWO OR MORE PARTS Section 74. The solution of a great many problems depends upon our being able to separate a number into 1 04 Fundamentals of High School Mathematics two or more parts. For example, if a man has a certain sum of money to invest, he may invest part of it in one thing, and part in another. The solution of such an ex- ample requires that we be able to divide a number into two or more parts algebraically. EXERCISE 75 PRACTICE IN DIVIDING A NUMBER INTO TWO OR MORE PARTS 1. The sum of two numbers is 20. (a) Express in algebraic form the second one if the first one is 12. (b) Express in algebraic form the second one if the first one is n. (c) Express in algebraic form the fact that the second one exceeds the first one by 4. 2. There are 36 pupils in a mathematics class. (a) Express algebraically the number of boys if there are 19 girls. (b) Express algebraically the number of girls if there are n boys. (c) State algebraically that there were 6 more girls than boys. 3. A farmer has two kinds of seed, clover seed and blue grass seed. If he has 100 lb. of both, express : (a) the number of pounds of clover seed if there were 24 lb. of blue grass seed ; {&) the number of pounds of clover seed if there were n lb. of blue grass seed ; Complete Solution of the Simple Equation 165 (c) the value of the clover seed {n lb.) at 20 cents per pound ; and the value of the blue grass seed at 15 cents per pound. (d) State by an equation that the value of both kinds together was $19. 4. Divide 20 into two parts such that the larger part exceeds the smaller part by 4. 5. A boy paid 18 cents for 20 stamps ; some cost two cents each and the remainder cost three cents. How many of each kind did he buy ? 6. During one afternoon a clerk at a soda fountain sold 200 drinks, for which he received $16. Some were 5 cents each; the others were 10 cents each. Find the number of each kind. 7. A grocer has two kinds of coffee, some selling at 30 cents per pound and some selling at 50 cents per pound. How many pounds of each kind must he use in a mixture of 100 pounds which he can sell for 31 cents per pound ? Section 75. Need for tabulating the data of word prob- lems. Many problems involve so many different state- ments that it is practically necessary to arrange the steps in the translation in very systematic tabular form. Take an example like this : John's age exceeds James* s by 20 years. In 15 years he will be twice as old as James. Find the age of each now. Before we can write this statement in the form of an equa- tion we must express in algebraic f oxm four different things : (1) John's age now ; (2) James's age now ; (3) John's age 1 66 Fundamentals of High School Mathematics in 15 years; and (4) James's age in 15 years. These four facts can best be stated in a table like this : (First step) Let it represent James's age now. (Second step) Tabulate the data : Table 9 Age now Age in 15 years Johns age n + 20 H + 20+15 James's age n n +i5 With all the facts expressed in letters we can now state the equation which tells the same thing as the original word statement ; namely : (Third step) n + 20 + 15 = \n + 15). We are now ready for the (Fourth step) the solution of the equation ; the steps are as follows : (1) n + 35 = 2^ + 30. (2) - n = -5. (3) ' » = 5. Therefore James's age now is 5, and John's age now is n + 20, or 25. (4) Check the accuracy of this result thus : In 15 years John will be 40 and James will be 20 ; or John will be twice as old as James, as the problem states. To be proficient in solving such problems, therefore, we first need practice in tabulating such facts as " age now" "age some other time," as in this example. Other types which involve the same need for tabulation will be taken up later. Complete Solution of the Simple Equation 167 PRACTICE IN REPRESENTING RELATIONS BETWEEN NUMBERS II. PROBLEMS RELATING TO AGE EXERCISE 76 1. A man is now 25 years of age. What expression will represent his age: {a) 10 years ago ? (c) x years ago ? (b) 8 years from now? (d) m years from now? 2. C is now n years of age. What expression will represent his age: {a) 12 years from now ? (c) y years ago? (b) 7 years ago? (d) m years from now? 3. A is now x years old. B's present age exceeds A's age by 8 years. What expression will repre- sent : (a) B's present age ? (b) the sum of their ages ? (c) the age of each 10 years ago? (d) the age of each 5 years from now ? {e) the sum of their ages in 5 years ? 4. A is now n years of age ; B is three times as old. Express algebraically : (a) B's present age ; (b) the age of each 4 years ago ; (c) the age of each 9 years from now. (d) State algebraically that B's age 4 years ago was 5 times A's age then. i6S Fundamentals of High School Mathematics 5. A's present age exceeds B's present age by 25 ( years. In 15 years he will be twice as old as B. Find their present ages. 6. C is six times as old as D. In 20 years C's age will be only twice D's age 20 years from now. What are their present ages ? 7. A man is now 45 years old and his son is 15. In how many years will he be twice as old as his son ? 8. A father is 9 times as old as his son. In 9 years he will be only 3 times as old. What is the age of each now ? 9. A's present age is twice B's present age ; 10 years ago A's age was three times B's age then. Find the age of each now. III. PROBLEMS BASED ON COINS Section 76. Another illustrative type of word problem which gives practice in tabulating data and thus in solving difficult word problems is the " coin problem." Take this example : Illustrative example. A man has 3 times as many dimes as quarters. How many of each has he if the value of both together is Sll? Here there are four distinct numbers to be expressed, as in the case of the age problem: (1) the number of quarters; (2) the number of dimes; (3) the value of the quarters in terms of a common base (for example, cents) ; (4) the value of the dimes in the same base (cents). The steps in the solution are clear, therefore, from the following illustrative solution : Complete Solution of the Simple Equation 169 (1) Let (2) Then n = the number of quarters. Table 10 Number Value quarters n 25 n dimes 3n 30 n (3) 25 n -f 30 n = 1100 cents. (4) .-. n = 20, number of quarters. (5) 3n = 60, number of dimes. (a) Value of the quarters = 85. (6) Fa/i/e of the dimes = 9 6. Total value = $ 11, as stated in the example. EXERCISE 77 PRACTICE IN EXPRESSING THE VALUE OF VARIOUS NUMBERS OF COINS 1. Express the value in cents of : (a) d dimes ; (d) 4 d half dollars ; (b) 3 d quarters ; (e) d dollars ; (c) 2 d nickels ; (/) of all the coins. 2. Express the value in cents of: (a) n nickels ; (c) (u + 5) quarters; (b) (3 - it) dimes ; (d) (12 - n) half dollars; {e) (30 — ri) nickels. 3. A purse was found which contained nickels and dimes, 20 in all. Find the number of each if the value of both was 11.60. 4. I received at a candy counter twice as many dimes as quarters, and 6 more nickels than dimes and quarters together. How many of each coin did I receive if the value of all was $7.50 ? 170 Fundamentals of High School Mathematics 5. A debt of $ 72 was paid with 5-dollar bills and --dollar bills, there being twice as many of the latter as of the former. Find the number of each kind of bill. 6. IS coins, dimes and quarters, amount to $2.25. Find the number of each kind of coin. 7. A cab driver received twice as many quarters as half dollars, and three times as many dimes as half dollars; in all he had $13. How many of each coin did he receive? IV. PROBLEMS BASED ON TIME, RATE, AND DISTANCE Section 77. In Chapter IV we saw that the motion of a train could be represented graphically. Now we shall learn how to solve this kind of problem by means of the equation. EXERCISE 78 PRACTICE IN SOLVING PROBLEMS BASED ON RELATIONS BETWEEN TIME, RATE, AND DISTANCE 1. Express the distance covered by an automobile in 10 hours if its rate is : (a) 18 miles per hour; (b) 5 miles per hour; (c) (r+ 3) miles per hour; (d) (2 x — 5) miles per hour. 2. A train runs for / hours. Express the distance it will cover at the rate of : {a) 85 miles per hour; (b) m miles per hour; (c) (r+ 6) miles per hour; (d) t miles per hour. Complete Solution of the Simple Equation 171 3. An automobile tourist sets out on a 400-mile trip. Express the time required if he goes at the rate of : (a) 40 miles per hour ; (b) 5 miles per hour ; (c) (r+ 10) miles per day; (d) (2 r — 3) miles per day. 4. How long will it require to make a trip of D miles at the rate of 15 miles per hour ? 5 miles per hour ? 5. At what rate must one travel to go D miles in 10 hours ? In t hours ? In / + 3 hours ? 6. A slow train travels at the rate of 5 miles per hour; a fast train travels 15 miles more per hour. Express : (a) the rate of the fast train ; (b) the distance passed over by each in 5 hours. (c) State algebraically that the two trains to- gether traveled 275 miles in 5 hours. 7. Two trains leave Chicago at the same time, one eastbound, the other westbound. The east- bound train travels 10 miles less per hour than the westbound train. Express : (a) the rate of each ; (b) the distance traveled by each in 4 hours. (c) Form an equation stating that they were 440 miles apart at the end of 4 hours. 8. Two trains, 350 miles apart, travel toward each other at the rate of 40 and 35 miles per hour, respectively. [72 Fundamentals of High School Mathematics (a) Express the distance traveled by each in / hours. (6) Form an equation stating the fact that the trains met in / hours. 9. Make formulas for d, for t, and for r, that can be used in any problem based upon uniform motion. 10. Illustrative example. Two bicyclists, 200 miles apart, travel toward each other at rates of 12 and 8 miles per hour respectively. In how many hours will they meet ? (1) Let t represent the number of hours until they meet. (2) Table 11 Time in hours Rate per hr. in miles Distance in miles For slow one For fast one t t 8 12 8t 12 1 (3) Then (4) 8 t + 12 t = 200. .-. * = 10. 11. Two men start from the same place, one going south and the other going north. One goes twice as fast as the other. In 5 hours they are 1^0 miles apart. Find the rate of each. 12. An eastbound train going 30 miles per hour left Chicago 3 hours before a westbound train going 36 miles per hour. In how many hours, after the westbound train left, will they be 519 miles apart ? Complete Solution of the Simple Equation 173 13. A bicyclist traveling 15 miles per hour was over- taken 8 hours after he started by an automobile which left the same starting point 4 J hours later. Find the rate of the automobile. 14. A starts from a certain place, traveling at the rate of 4 miles per hour. Five hours later B starts from the same place and travels in the same direction at the rate of 6 miles per hour. In how many hours will B overtake A ? V. PROBLEMS INVOLVING PER CENTS Section 78. Many problems involving per cents may be solved by algebraic methods. EXERCISE 79 PRACTICE IN SOLVING PERCENTAGE PROBLEMS 1. What does 10% mean ? 5% ? r%J 2. Indicate 4 % of $600 ; 5 % of 1275. 3. Express decimally 5 % of / ; 8 % of c ; 6^ % of b. 4. A man paid c dollars for an article. He sold it at a gain of 25 %. Express : (a) the gain in dollars ; (b) the selling price. (c) State algebraically that he sold the article for $2.50. 5. A merchant sold a suit for $25, thereby gaining 25%. If the cost is represented by c dollars, what will represent : (a) the gain in dollars ? (b) the selling price in terms of c ? (c) State algebraically that the selling price was $25. 174 Fundamentals of High School Mathematics 6. Solve each of the following equations: (*) .20 x= 180 (b) x + Mx = 3.18 (c) c + .10c = VX r > (d) vi - Xhm = 21.25 (e) p + .04/ = 520 if) x- .50 x = 18.75 \g) 2 - Sx- ,5x= 7 (h) 1.15 x- \x= 1000 7. Find the cost of an article sold for $ 156 if the gain was 10 %. (Use c for the cost.) 8. What number increased by 66| % of itself equals 150 ? 9. After deducting 15 % from the marked price of a table, a dealer sold it for % 21.25. What was the marked price ? 10. A dealer made a profit of $ 3690 this year. This is 18 % less than his profit last year. Find his profit last year. 11. A number increased by 12.5 % of itself equals 213. What is the number ? 12. A shoe dealer wishes to make 25 % on shoes. At what price must he buy them in order to sell them at $4.50 per pair? 13. A furniture dealer was forced to sell some dam- aged goods at 14 % less than cost, and sold them for $ 129. How much did they cost ? 14. A man sold a suit of clothes for % 30.25. What per cent did he gain if the clothes cost him $ 25 ? Complete Solution of the Simple Equation 175 VI. INTEREST PROBLEMS Section 79. Many interest problems can be more easily solved by algebraic equations than by the methods of arithmetic. EXERCISE 80 1. Express the interest on $150 at 5 % for 1 year; for 3 years ; for / years. 2. Express the interest on P dollars at 6 % for 1 year ; for 3 years ; for / years. 3. Express the simple interest on #500 for 1 year at r °/o ; for 4 years. 4. A man borrowed a certain sum of money at 6 %. Express : (a) the interest for 2 years. (b) State algebraically that the interest for three years was $48. 5. What principal must be invested at 6 % to yield an annual income of $ 57 ? 6. For how many years must $ 2800 be invested at 7 % simple interest to yield $ 833 interest ? 7. What is the interest on P dollars at r% for / years ? 8. A man invests part of 1 1000 at 4 %, and the remainder at 6 %. If x represents the number of dollars invested at 4 %, express : (a) the annual income on the 4 °j investment; (b) the amount of the 6 % investment ; (c) the annual income on the 6 % investment. (d) State algebraically that the annual income on the 4 % investment exceeds the annual income on the 6 °/ investment by $ 20. 176 Fundamentals of High School Mathematics 9. Part of $1200 is invested at 5% and the re- mainder at 7 %. The total annual income from the two investments is $67. What was the amount of each investment ? 10. Ten thousand dollars' worth of Liberty Bonds yield an annual interest of $ 370. Some pay o\ %, and the remainder pay 4 %. Find the amount of each kind of bond. 11. A 5 % investment yields annually $5 less than a 4 % investment. Find the amount of each investment if the sum of both is $ 800. VII. PROBLEMS CONCERNING PERIMETERS AND AREAS Section 80. The following examples are based on squares and rectangles. EXERCISE 81 1. The length of a rectangle exceeds twice its width by 12 in. Represent its width by w. (a) Make a drawing to represent it. (b) Express its length. (c) Express its area. (d) Express its perimeter. (e) State that its perimeter is 84 in. 2. The length of a rectangle is 9 in. more, and the width is 6 in. less, than the side of a square. (a) Make a drawing for each. (b) Express the dimensions of the square. (c) Express the dimensions of the rectangle. (d) Express the perimeter of the rectangle. (c) Express the area of the rectangle. (/) State algebraically that the sum of the perim- eters is 168 in. Complete Solution of the Simple Equation 177 3. The base of a triangle exceeds its height by 10 inches. {a) Make a drawing for the figure. (b) Express its base and height. (e) Express its area. (d) State that its area is equal to the area of a rectangle whose dimensions are 8 in. and 5 in. 4. The length of a rectangle is 4 feet more, and its width is 2 feet less, than a square whose perim- eter is P inches. Express : (a) the side of the square ; (b) the dimensions of the rectangle; (c) the perimeter of the rectangle. (d) Find the value of P if the perimeter of the rectangle is 44 inches. VIII. PROBLEMS BASED ON LEVERS Section 81. A teeter board is one form of lever. The point on which the board rests or turns is the fulcrum; the parts of the board to the right of and to the left of the fulcrum are the lever arms. If a boy at A, who just balances a boy at B, moves to the left while B remains stationary, it is clear that the left side goes down. But if the boy at B moves closer to the 178 Fundamentals of High School Mathematics fulcrum while A remains stationary, then A goes down. It is also clear that boys of unequal weight cannot teeter unless the heavier boy sits closer to the fulcrum. There is a mathematical relation between the weight on the lever arm and its distance from the fulcrum. Two boys will balance each other when the weight of one times his dis- tance from the fulcrum is equal to the weight of the other times his distance, or in general, when weight times distance on one side equals weight times distance on the other side. This law or relation may be tested by placing equal coins at different positions on a stiff ruler balanced on the edge of a desk. Try this experiment. See whether 2 pennies placed 4 inches from the fulcrum (at the center of the lever) will balance 1 penny placed 8 inches from the ful- crum. See whether 6 pennies placed 2 inches from the fulcrum will balance 3 pennies placed 4 inches from the fulcrum. Thus, to make a lever balance, the product of weight and distance from the fulcrum on one side must equal the product of weight and distance from the fulcrum on the other side. EXERCISE 82 Problems based on levers. Make a drawing for each. 1. John weighs 80 lb. and sits 4 ft. from the ful- crum. Where must Robert sit if he weighs 90 lb. ? 2. A, weighing 120 lb., sits 4* ft. from the fulcrum, and balances B, who sits 5 ft. from the fulcrum. What is B's weight ? Complete Solution of the Simple Equation 179 3. A hunter wishes to carry home two pieces of meat, one weighing 40 lb. and the other 60 lb. He puts them on the ends of a stick 4 ft. long and places the stick across his shoulder. Where must the fulcrum (his shoulder) be placed to make the weights balance ? 4. Two children play teeter, one on each end of a board 9 ft. long. Where must the fulcrum be if the children weigh 60 and 80 lb. respectively ? 5. Could three children teeter on the same board ? How ? 6. A and B sit on the side of the fulcrum. A weighs 100 lb. and sits 5 ft. from the fulcrum ; B weighs 80 lb. and sits 3 ft. from the fulcrum. Where must C sit to balance the other two, if he weighs 150 lb. ? 7. Show that — \ = _2 is the formula or law for a balance on a teeter board. Are the weight and distance directly proportional ? SUMMARY This chapter has taught all the steps involved in solv- ing a simple equation : 1. Removal of parentheses. 2. Getting rid of fractions. 3. Collecting terms on each side. 4. Getting rid of known terms on one side and un- known terms on the other side. 5. Dividing each side by the coefficient of the un- known. 1S0 Fundamentals of High School Mathematics 6. Checking by substituting the obtained value of the unknown in the original equation. Many kinds of word problems have been solved. Tabu- lating the information or data of such problems is a great help in solving them. A systematic method always pays big dividends in any kind of work. EXERCISE 83 MISCELLANEOUS PROBLEMS 1. A grocer has two kinds of tea, — some worth 60 f! per pound and some worth 75 per pound. He has 20 lb. more of the former than of the latter kind. How many pounds of each kind has he, if the value of both kinds is $45.75 ? 2. I bought 45 stamps for $1.05. If part of them were 2-cent stamps and part 3-cent stamps, how many of each did I buy ? 3. The sum of the third, the fourth, and the eighth parts of a number is 17. What is the number? 4. John has I as many marbles as Harry. If John buys 120 and Harry loses 23, John will then have 7 more than Harry. How many has each boy ? 5. A clerk spends \ of his yearly salary for board and room, \ for clothes, £ for other expenses, and saves $880. What are his annual expenses? 6. A father left one third of his property to his wife, one fifth to each of his three children, and the remainder, which was $1200, to other rela- tives. Find the value of his estate. Complete Solution of the Simple Equation 181 7. Ten years ago A was one third as old as he is at present. Find his age now. 8. Evaluate the formula C='-± — • ' if 7^=20. 9. A merchant bought goods for §500, and sold them at a gain of 5%. What was the selling price ? 10. If in problem 9 the merchant had sold the goods at a gain of x per cent, what would have been the selling price ? 11. A 6-foot pole casts a shadow 4 J ft. in length. At the same time how long is the shadow of an 8-foot pole ? 12. The ratio of two numbers is f. Find each number if their sum is 56. 13. Two numbers differ by 70 ; the ratio of the larger to the smaller is -|-. Find each number. Cj 14. In Fig. 97, ZC= 90°, A A = 37°, and AC=2±. Find AB, BC, and Z B. Fig. 97 15. In general, which is the larger, the sine or the tangent of an angle ? Show by a drawing. 16. The highest office building in the world (the Woolworth Building, New York City) casts a shadow 1240 ft. long at the same time that a boy 5 ft. tall casts a shadow 8 ft. long. What is the height of the building ? [82 Fundamentals of High School Mathematics 17. The table below gives the annual cost of pre- mium per $1000 life insurance, at various ages. Age 21 25 30 35 40 45 50 Premium 18.25 20.04 22.60 26.40 30.50 3610 45.20 Show this graphically. Measure age along the horizontal axis. CHAPTER XI HOW TO SOLVE EQUATIONS WHICH CONTAIN TWO UNKNOWNS I. GRAPHICAL SOLUTION Section 82. Importance of skill in drawing the picture or graph of an equation. In Chapter IV we learned how to represent and to determine the relationship between quantities that change together. Three methods of doing this were studied : (1) the tabular method ; (2) the graphic method; (3) the equational or formula method. One of the most important facts for us to recall is that the graph and the equation tell exactly the same tiling. For exam- ple, on 'page 51, the line BC and the equation C= .12n tell exactly the same thing. Any information that you get from the equation you can also get from the graph. Furthermore, relationships can be seen more easily from graplis than from tables or equations. For these reasons, and since much of our later work in mathematics makes use of graphic methods, we need to be skillful in drawing the line which stands for an equation. Section 83. We need to know how to locate or to "plot " points.. But a line may be regarded as a series of points. Thus, to represent or locate a line we have to locate a series of its points. It happens that much of our elemen- tary work, furthermore, deals with straight lines. This kind of line, clearly, can be fully determined by locating any two of its points. HOW TO LOCATE OR PLOT POINTS Thus, we see that the important thing in "graphing" is how to locate, or to represent, points. In every graph that you have already constructed you have had to locate 183 184 Fundamentals of High School Mathematics joints through which to draw the line. For example, in constructing a cost graph it is necessary to locate several points representing the cost of different numbers of units of the article. Let us study more carefully how points are located. Section 84. How points are located on maps. (1) Points are located on maps by means of latitude and longitude. Any point on the earth's surface is definitely located by stating its distance cast or west of the prime meridian, and its distance north or south of the equator. Thus, to the nearest degree, the location of New York is 74° W. and 41° N. because it is 74° west of the prime meridian and 41° north of the equator. Similarly, the position of Chicago is 88° W. and 42° N. ; that of Paris, 2° E. and 49° N. (2) This same method is used by many cities in number- ing their houses. Two streets, which make right angles with each other, are selected as reference streets. Any house or building is completely located, then, by stating the number of blocks it is east or west, and north or south, of these reference streets. Section 85. How points are located on drawings. By a method similar to that above, we locate points on 'paper. Instead of using the equator and the prime meridian as our reference lines, we take two lines, — for convenience, one horizontal and the other vertical, — which make a right angle with each other. Any point may be located, then, by stating its distance to the fight of, or to the left of, the vertical reference line ; and its distance above or below the horizontal reference line. Thus, in Fig. 98, point A is 2 units to the right of, and 1 unit above, the reference lines ; point B is 2 units to the left of, and 3 units below, the reference lines ; point C is 8 units to the left of, and 2 units below, the reference Solving Equations with Two Unknowns 185 -x D A c B X -Y Fig. 98 lines ; and point D is units to the right or left of, and 3 units above, the reference lines. Section 86. The point from which distances are meas- ured : The origin. The point in which the two axes meet, or their intersection point, is called the origin. It is the point from which we measure distances, either way. The origin is usually lettered with a capital O, as in Fig. 98. Section 87. How distances are distinguished from each other. It would be laborious to state that a particular point is "to the right of" or "to the left of "some refer- ence line, each time we refer to it. To avoid this, it has been agreed to call distances to the right of the Y-axis positive, and distances to the left of the Y-axis negative. Similarly, distances above the X-axis are positive, and dis- tances below the X-axis are negative. It is very important 1 86 Fundamentals of High School Mathematics to remember these facts because we use them so frequently in graphic work. Thus, in Fig. 98, the position of point A is described by the numbers + 2 and +1, or by (2, 1). This means that point A is 2 units to the right of the F-axis, and 1 unit above the .Y-axis. Similarly, the position or location of point B is described by the numbers — 2 and — 3, or by ( — 2, — 3); this means that point B is 2 units to the left of the F-axis and 2 units above the ^Y-axis. In the same way, the position of point C is described by the numbers — 3 and — 2, or (— 3, — 2); this means that point C is 3 units to the left of the F-axis and 2 units below the JYaxis. At this time the student should note that in stating the location of a point, its distance to the right of, or to the left of, the F-axis is always given before its distance above or below the X-axis. This is done to avoid confusion. That is, the x-distance is always first, the ^/-distance second. Section 88. Plotting a point. By "plotting a point" we mean the locating, on cross-section paper, of a point whose x -distance and j/-distance are known. Thus, to plot A, whose ^-distance is -f 3 and whose ^-dis- tance is + 4, usually written (3, 4), means to locate on the cross- section paper a point jyr 3 units to the right of, and 4 units above, the origin, as in Fig. 99. In the same way, the point ( — 2, 1) is point B on the graph. +Y ~lr >" JJ3_ 1r .4; -X\ •r. ■X Fig. 99 Solving Equations with Two Unknowns 187 EXERCISE S4 PRACTICE IN PLOTTING POINTS 1. The ^--distance of a point is + 3, i.e. it is 3 units to the right of the vertical axis. Is it definitely located ? Why ? 2. The j'-distance of a point is — 4, i.e. it is 4 units below the horizontal axis. Is it definitely located? Why? 3. A certain point is on both axes. What are its x- and j-distances ? 4. Plot the points whose position is determined by the following : (4, 2), (5, 6), (- 3, 2), ( - 4, - 1), and (-6, 2). 5. Plot the following: (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (8, 2), (10, 0). 6. Plot the following : (12, - 2), (15, - 5), (18, - 8), (10, 0), (5, 5). 7. Plot the following: (21 3), (If, 5), (-21, 3). HOW TO DRAW THE GRAPH OF AN EQUATION WHICH CONTAINS TWO UNKNOWNS Section 89. The picture of an equation. Now that we have learned how to locate, or plot, points, we come to the main purpose of the chapter : to shozv Jiozv equations can be solved grapliically. First illustrative example. Let us take an equation which contains two unknowns, such as, In this equation the value of y changes as the value of x changes. Clearly, the value of y depends upon the value of x. For example, if x = 1, then y — 5 ; if x = 2, then y = 7, etc. A table will help to show this relation between the unknowns, x and y. [88 Fundamentals of High School Mathematics Table 12 If x equals 1 2 3 4 5 O -1 -2 -3 -4 -5 then y equals 5 7 9 11 13 3 1 -1 -3 -5 -7 If we select any particular value of x, and the corresponding value of y which accompanies it, such as 1 and 5, or 2 and 7, we may think of them as completely describing the position of points on a graph. Thus, (1, 5), (2, 7), (3, 9), etc., definitely locate the position of the points. Plotting these points with respect to an X- and F-axis, we get a series of points, such as Fig. 100. By joining these points we obtain a straight line, which is the picture or the graphical representation of the equation y — 2x + 3. +Y -x- *-r- T 22 ±tt -e07 Z^ f -Z+ A v 7 T I t / T / _4T ta Ti / J t / t / f 4 7 -Y Fig. 100 Solving Equations with Two Unknowns 189 Second illustrative example. For a second example, let us consider an equation in two unknowns which shows that the sum of two numbers is always 10, such as x -f y = 10. It is clear that one number, x, might be 2, and if so, that y must be 8 ; or that x might be 4, and if so, the other number, y, must be 6. Thus the two unknowns, x and y, may have many different values. A table helps to show this. Table 13 If x equals l 2 3 4- O -1 -2 -3 then y equals 9 8 7 6 10 11 12 13 Now we may think of any pair of related numbers, such as 1 and 9, or 2 and 8, as describing the position of points on this line. Thus, (1, 9), (2, 8), (3, 7), (4, 6), etc., show the location of points on the graph. Plotting these points, we have Fig. 101. By joining these points we obtain in this illustra- tive example the "graph"' or picture of the equation x + y = 10. Expressed in another way, we have represented graphically the relation between two numbers whose sum is always 10. +Y -X- s__ _^,_ s s -V^ v jL - 5^ ^^ S3I sL Ia -O-'' n -\ -fc V _\ <v7 -Y Fig. 101 joo Fundamentals of High School Mathematics EXERCISE 85 PRACTICE IN REPRESENTING GRAPHICALLY EQUATIONS WHICH CONTAIN TWO UNKNOWNS 1. In the equation y = 2x + 5, find the value of y when x == 1 ; when x = 2 ; when x = 4 ; when x =s ; when x = — 3. Make a table similar to the one above, showing these related pairs of numbers. 2. In the equation x = y — 4, find the value of x when y = ; when y — 1 ; when j = 4 ; when j = 6 ; when j/ = — 2 ; when y = — 6. Tabu- late. 3. Plot the equation given in Example 2. 4. Plot the equation x + y = 6. Hint : First tabu- late related values of x and j/. Use only four points. 5. Plot, or graph, y = 5 + x. 6. Plot 2;r+ y = 6, orj = 6 - 2*. 7. Graph x — 2 j/ = 5, or x = 2y + 5. 8. Plot 3;r — y = 8, or j = 8.r — 8. Section 90. An easier method of plotting a line. A straight line is definitely determined or located if any two of its points are known. If these points are not too close together, they fix the plotted position of the line just as accurately as eight or ten points. Therefore, in plotting a straight line, it is sufficient to plot only two points, unless they are quite close together. The easiest points to plot are those on the axes ; that is, the points where the line cuts the .r-axis and the jj/-axis. By referring to Fig. 101, or to the graph of any line, you Solving Equations with Two Unknowns 191 will see that the x-distance of the point in which the line acts the vertical, or y-axis, is alzvays 0, and that the y-dis- tance of tJie point in which the line acts the horizontal, or x-axis, is alzvays 0. Thus, if we let x be in any equa- tion, such as 2x—y = 6, we find the point in which the line cuts the j/-axis. If x is in 2x—y = 6, we see that y equals — 6, which shows that the line cuts the y-axis at a point (0, —6); that is, 6 units below the origin. In the same way, if we let y be in any equation, we find the point in which the line cuts the ^r-axis. In this particular equation, 2x—y = G, if y is 0, then x is 3, which shows the point in which the line 2x — y= 6 cuts the ^r-axis. This shorter method requires only the following brief table : Table 14 X equals ? y equals ? EXERCISE 86 1. If x = 0, what is y in the equation 2 x +y = 8 ? What is x if y = ? From these two sets of values for x and y, plot the equation. 2. Given 4,r — 2j/ = 8. Plot by finding where the line cuts the axes. 3. Where does the graph of 5x -\-2y = 10 cut the jr-axis ? thej/-axis? 4. Where does the graph of 2x — 3y = — 6 cut the ;r-axis ? the j/-axis ? Plot. 5. Gr aph 2\ vX + y = 5. ig2 Fundamentals of High School Mathematics HOW TO SOLVE GRAPHICALLY EQUATIONS WITH TWO UNKNOWNS Section 91. When is an equation with two unknowns solved? In equations with only one unknown, such as 3 x + 4 = 19, we found that there was only one value for x which would satisfy the equation; namely, x = b. If we substitute 5 for x in this equation, giving 15 + 4 = 19, we find that 5 satisfies the equation. Any other number sub- stituted for x would not " satisfy the equation." But now consider an equation which has two unknowns, such as ;r+j=8. Here we see that x might be 3 and y would be 5 ; or x might be 6 and y would be 2 ; or x might be 10 and y would be — 2. Thus, there are a great many sets of values of x and y which could satisfy the equation x+y = S. This will be made clear as you work the fol- lowing examples. EXERCISE 87 1. Give four sets of values of x and y that will satisfy the equation x—y= 6. 2. Will x = 4} 2 and y = 3 satisfy the equation \x — y = 15 ? Does x = 5 and j/=4 satisfy it ? 3. If the equation x+y = 8 is plotted, would the points (5, 3) lie on the line representing the equation ? (3, 4) ? (1 0, - 2) ? (1, 7) ? 4. Does the graph of the equation 2x — y= 7 pass through the point (5, 3) ? (4, 2) ? Solving Equations with Two Unknowns 193 We have shown that an equation with two unknowns is solved when a set of values for the unknowns is found which satisfies the equation. Section 92. Linear equations. The fact that the graph of an equation which contains two unknowns, each of the first degree {i.e. no squares or cubes), is always a straight line, has led to the lame linear equations. Thus, 2x +y = 5, x + 5 = 10, etc., are linear equations. TWO LINEAR EQUATIONS MAY BE EASILY SOLVED BY PLOTTING THEM ON THE SAME AXES Section 93. It is a very common problem in mathe- matics to have to find one set of values which will satisfy each of two equations having two unknowns. For ex- ample, what single set of values will satisfy each of these equations ? \x+y = -J' = ~ It is clear that x = 6 and y = 2 or (6, 2) will satisfy the first equation, but not the second one ; in the same w T ay x = i and 7 = 4 or (4, 4) satisfies the first equation, but not the second one ; x = 6 and y = 5 satisfies the second equation, but not the first one. OUR PROBLEM IS TO FIND ONE SET OF VALUES THAT WILL SATISFY BOTH EQUATIONS This can be done, graphically, by plotting both equations on the same axes, because in that way we can find a point common to the two lines ; that is, the point in which two lines intersect. The coordinates of this point will satisfy both equations. Figure 102, on the following page, shows both equations plotted on the same axes. Note that the lq4 Fundamentals of High School Mathematics * - 'V -5> ■ JL-L - -^ -At K - zl ^r-tS-' 5?7 *cr /^ r s r j S _ _. -fi- -/ v»fc ^ 7 Sp* r S*^ I Kxl- -T 5Z 7 S r^ / _/ Fig. 102 two lines intersect at the point (5, 3). This point of inter- section of the two lines gives a single set of values, x = 5 and y — 3, which satisfies both equations. (Show that x = 5 andj= 3 checks for each of the equations.) EXERCISE Find a set of values that will satisfy each of the follow- ing pairs of equations, by finding the intersection point of their graphs : 1. 2. 4. 5. x + y = 6 2x-y = 3 y=2x+Z y = x+l \x — y = 5 \2x+y = 7 x _ 3 j, = _ 2 j/ = .r + 6 6. 9. \ a -2b = -5 x+y= 10 3.r + 3j/ = 6 „v = j/ + 6 _r + j = 1 x — y = 5 Solving Equations with Two Unknowns 195 +Y ->r- ^ ^,- ST s± : ,_ In V St JS ^ 5 1 v "^ V- ^ X T ■ s K .v. • ^t Xii- --X- . . -4.TT A ISji _ \ I A ±t - : s££ ^ K ^ 1 N Sr s ^ fr '» '5? s V> ^ or. ^ "x^ ^ - ^V 1 — 1 — 1— J — 1 1 1 1 1 1 1 -Y Fig. 103 Section 94. Equations whose graphs are parallel lines, i.e. inconsistent equations. Figure 103 shows the graphs of the equations represented below. (1) \x+y = ± (2) \2x + 2y = -6 Note that the lines do not intersect, but are parallel. What single set of values of x and y will satisfy each of these equations ? Evidently there is none, for they have no point in common. Such equations are generally called incon- sistent, to distinguish them from the kind that are satisfied by some set of values. The latter kind, those whose graphs intersect, are often called simultaneous equations. 196 Fundamentals of High School Mathematics EXERCISE 89 Graph each of the following pairs of equations to determine which pairs are inconsistent and which are simultaneous : y — *==4 3 jx+y = 4; { x — 6 = y \ x — y = 6 l2j/ + 10 = 4.r ' \x + y=12 SUMMARY This chapter should make it clear that : 1. An equation may be plotted or graphed by locat- ing a series of points, the x and y values of which will satisfy the equation. 2. Two equations are solved graphically by finding the x and y values of the point of intersection of their lines. REVIEW EXERCISE 90 1. From the sum of — 6 and + 10 take — 8. 2. The product of two numbers is — 40 y; one of them is + 10. What is the other ? 3. State the four principles, or axioms, used in solv- ing equations. Illustrate in solving the equation by -8 =+2y- 50. 4. If A = 4:X + Sy and B = ix — 3y, what does A + B equal ? What does A - B equal ? 5. Does -- = '-^— ? Does - = — ? Does - = — ? 7 7-2 b be 2 10 State the principle involved in these examples. Solving Equations with Two Unknowns 197 Age Boys who leave school at the age of 14 earn weekly wages as indicated Boys who leave school at the age of 18 earn weekly wages as indicated 14 $ 4.00 16 5.00 18 7.00 f 10.00 20 9.50 15.00 22 11.00 20.00 24 12.00 24.00 25 13.00 30.00 6. Studies have been made to determine the money value of a high school education. The table above shows the average weekly earnings for boys who leave school at the age of 14, and for those who remain in school until they are 18 years old. Graph the earnings for each class of boys on the same axes. Measure age along the horizontal axis. Interpret the graph. If a boy knew that he would live to be only 25 years old, would it pay him, in dollars, to go to high school ? How much ? 7. The areas of two circles are directly proportional to the squares of their radii, Compare their areas if the radius of one circle is 3 times the radius of the other. 8. In solving a particular problem, how do you tell whether to use the sine, cosine, or tangent? Illustrate by specific examples. CHAPTER XII HOW TO SOLVE EQUATIONS WITH TWO UNKNOWNS (Continued) II. SOLUTION BY ELIMINATING ONE UNKNOWN Section 95. The need for a shorter method of solving equations with two unknowns. In the previous chapter we saw that equations with two unknowns can be solved by graphic methods. The exclusive use of that method, how- ever, would require a great deal of time, and would necessitate that we have cross-section paper at all times. Fortunately, there is a sJwrter method which can be used. This is a method by which we eliminate one of the unknowns. By eliminating or getting rid of one of the unknowns, we obtain an equation with only one unknown. The following illustrative examples will explain the differ- ent ways by which one of the unknowns is eliminated. This chapter will show three methods of elimination. I. ELIMINATION BY COMPARISON; THAT IS, BY EQUATING VALUES OF ONE OF THE UNKNOWNS Section 96. Equating values of one of the unknowns. This method is illustrated by the following example: First illustrative example. Find the value of x and of y in the following equations : f*+ 2/ = 10, jx-3y = -6. Solution : (1) (2) Solving for x in equation (1), x = 10 - y (3) and in equation (2), x = 3 y - 6. (4) Comparing or equating the values of x, 3 y- 6 = 10- y, (5) which gives 4 y = 16, # or y = 4. (6) (7) Solving Equations with Two Unknowns 199 Substituting 4 for y in (1) or (2) gives jc = 6. Checking in equations (1) and (2), 6 + 4 = 10. 6 -12 ==-6. Elimination by comparison is based upon a fact which was illustrated in the graphical solution of equations with two unknowns ; namely, that at the point of intersection of the two lines, the value of x in one equation is the same as the value of x in the other equation, and the value of y in one equation is the same as the value of y in the other. For this reason, we may form an equation by equating, or placing equal to each other, the two values of x y which were 10— y and 3j — 6. This process gets rid of, or eliminates, the unknown x and gives us an equation with only one unknown, y. Second illustration of the method of eliminating one unknown. The same results could have been obtained by finding the value of y in each of the two given equations : jc+ y = 10. (1) x-Zy=-Q. (2) From (1), y = l0-x. (3) From (2), -3jf=— 6-x, (4) Comparing the values of y 10 — x = ^ « (6) 3 Multiplying by 3, 30 - 3 x = 6 + x, (7) or -4x=-*, (8) or x = S. (9) Substituting in (1) or in (2), y = 4. (10) 200 Fundamentals of High School Mathematics EXERCISE 91 PRACTICE IN ELIMINATING ONE OF THE UNKNOWNS BY COMPARING, OR EQUATING, ITS VALUES AS OBTAINED FROM THE TWO EQUATIONS Solve and check each of the following : * = 2^-3 f.r-3/ = \x= by -21 4 * \±s+t=l± fr-j/=10 5 <2b + 3c = 6 j,r = 16-2j/ ' \b = 5c + l6 [y + 2x=12 ( X + E>j/ = 1 3 * [5^ + ^ = 42 6 * |2.r+6j/ = -2. 7. The sum of two numbers is 14; the larger ex- ceeds the smaller by 2. Find each number. 8. Twenty coins, dimes and nickels, have a value of $1.70. Find the number of each. 9. A boy earns $ 2 per day more than his sister; the boy worked 8 days and the girl worked 6 days. Both together earned $44. What did each earn per day ? 10. 11. 1^ + ^=10 12 * \§ x = u-y 2;r + 3j = 5 [j/ = 2.r-10 Zx-y=2 13 * [x=2y-U II. ELIMINATION BY SUBSTITUTION; THAT IS, BY SUB- STITUTING THE VALUE OF X FROM ONE EQUATION IX THE OTHER EQUATION Section 97. This method of elimination will be illus- trated by working the same problem which we used in the previous section. Solving Equations with Two Unknowns 201 Illustrative example. Find the value of x and of y in the following equations : f*+ $(=10, (1) l*-3y = -6. (2) Solving equation (1) for x, we get x = 10 - y. (3) Substituting 10 — z/ for x in (2) gives 10-y-3y = -6, (4) or _4y=-16, (5) or y = 4. (6) Substituting 4 for y in (1) or (2) gives x = 6. (7) Here, as when we eliminate one unknown by " compari- son," our real aim is to get an equation which contains only one unknown. We found from equation (1) that x = 10 — y. This value of x must be true for both equations. (Recall that x is the same for both equations, or for both lines, at their point of intersection.} For this reason we may sub- stitute 10— y in place of x in the second equation. This gives an equation in one unknown ; namely, y. The same results could have been obtained by finding the value of y, instead of the value of x, from one of the equations and substituting it in the other equation. For example : Second illustration of the method of eliminating by sub- stitution. Jx+ y = io. (l) U-3z/=-6. (2) From equation (1), y = 10 — x. (3) Substituting 10 — x for y in (2) x _3(10_x) = - 6, (4) or x - 30 + 3 x = - 6, (5) or 4x = 24, (6) or x = 6, (7) and y — 4, as before. 202 Fundamentals of High School Mathematics EXERCISE 92 Solve by the method of substitution and check each result : fr-2j=10 \y = x ' \ Zx + 2y= 6 ' }3.r + 4j/=7 Lr-3j=-l f^-+j = 8 )4.r-j = -15 \2^— j/ = 10 2. 7. The sum of two numbers is 102 ; the greater exceeds the smaller by 6. Find the numbers. 8. The difference between two numbers is 14, and their sum is 66. Find the numbers. 9. 12 coins, nickels and dimes, amount to $1.05. Find the number of each kind of coin. 10. The perimeter of a rectangle is 158 inches ; the length is 4 feet more than twice the width. Find the dimensions of the rectangle. 11. Bacon costs 10 cents per pound more than steak. Find the cost per pound of each if 4 pounds of bacon and 7 pounds of steak to- gether cost $3.48. 12. A part of $4000 is invested at 4% and the re- mainder at 5%. The annual income on both investments is $185. Find the amount of each investment. 13. The quotient of two numbers is 2, and the larger exceeds the smaller by 7. Find the numbers. 14. Oranges cost 20 cents per dozen more than apples. A customer bought 10 dozen oranges Solving Equations with Two Unknowns 203 and 4 dozen apples and received 20 cents in change from a 5-dollar bill. Find the price per dozen of each. 15. 16. 17. lf/ = <7 — 4 y 2 2£_3j/_7 X 4 ~~3 18. III. ELIMINATION OF ONE UNKNOWN BY ADDING OR BY SUBTRACTING THE EQUATIONS Section 98. A great many equations in two unknowns can be most easily solved by this method. The following example illustrates it. Illustrative example. Find the value of x and y in these equations : (2x-y = 5, (1) { x + y = \Z. (2) Adding equation (1) and equation (2) gives 3jc = 18, (3) or x = 6. (4) Substituting 6 for x in (1) and (2) gives y = 1. (5) Check: Substituting 6 for x and 7 for y in (1) and (2) gives 12-7 = 5. ^ (6) 6 + 7 = 13. (7) It happens in this example that one of the unknowns, y t is eliminated by adding the corresponding members of the given equations. In many examples it is possible to eliminate one of the unknowns by subtracting the mem- bers of one equation from the corresponding members of 204 Fundamentals of High School Mathematics the other. In many other examples, however, it is im- possible to eliminate one of the unknowns directly, either by adding or by subtracting the members of the two equations. For illustration, take this set of equations : Illustrative example. f*-2y = 8, (i) sx + y = e, (2) If we add the corresponding members of the two equa- tions, we get the equation 3x — y = 14. But this does not eliminate either of the unknowns. In the same way, if we subtract (2) from (1), we get the equation — x — 3 y = 2. Again, this does not eliminate either one of the unknowns. This shows that addition or subtraction of the members to the equations will not eliminate one of the unknowns unless one of them, the x or the y> has the same coefficient in both equations. Now let us make y in the second equation have the same coeffi- cient as y in the first equation. To do so, the second equation must be multiplied through by 2. This gives f4x + 2y = 12, (3) I *-2y = 8. (1) Now, by adding (3) and (1), we get rid of y, obtaining : 5 x = 20, (4) or x = 4. (5) Substituting in (1) or (2), y = - 2. (6) Check : 16 - 4 = 12. (7) 4 + 4=8. (8) An important question naturally arises here : When do we eliminate by addition and when by subtraction ? This can be answered by referring to an example. \x+y=U, (1) 2x+y = 4. (2) Solving Equations with Two Unknowns 205 Would either x or y be eliminated by adding the corre- sponding members of these equations ? Certainly not, for that would give Sx + 2y = 15. Now, would either x or y be eliminated by subtracting the members of one equation from those of the other ? Yes, for we should have — x = 7. However, if the second equation (2) above had been 1x — y — 4, then we should eliminate jj/ by adding {1)2li\& (2). From these examples we come to the following conclu- sions about eliminating one of the variables : I. If the variable we wish to eliminate has the same sign in both equations, then it is elim- inated by subtracting the members of one equa- tion from the members of the other equation. II. If the variable we wish to eliminate has dif- ferent signs in the two equations, it is elim- inated by adding the corresponding members of the equations. III. No variable can be eliminated either by addi- tion or by subtraction unless it has the same co- efficient in both equations. If it does not have the same coefficient in both equations, then we must multiply one, or both, of the equations by such a number, or numbers, as will make that variable have the same coefficient. Thus, to eliminate x in the following equations : J2*- y = 8, (l) \3x + 4z/ = 23. (2) It is necessary to multiply (1) by 3 and to multiply (2) by 2. This gives the following equations : f6x-3y = 34, (3) \6x + 8y = 46. (4) Now the variable x can be eliminated by subtracting (4) from (3), which gives -Hi/ --22, or y = 2, 2o6 Fundamentals of High School Mathematics EXERCISE 93 ELIMINATION BY ADDITION OR SUBTRACTION Solve and check each of the following : lx+y = b [2>a-b = -2 • \x-y = 2 [4tf + £ = -12 l2x+3y = U I2x-j = ll 2 - \3.r-3j = l 6- \x- 3^ = 13 I2r+s = 9 i3b + 2c = 5 3 - \r-s = 12 T \2& + c = 3 fr-j/ = -8 (4jr-3y = 8 4 I 8 I ' \ x +y = -4: * \x-Ty = 2 9. Find two numbers whose sum is 100 and whose difference is 18. 10. In an election of 642 votes an amendment was carried by a majority of 60 votes. How many voted yes and how many no ? 11. The admission to a school play was 25 cents for adults and 15 cents for children. The proceeds from 267 tickets were $ 50.05. How many tickets of each kind were sold ? 12. A purse containing 18 coins, dimes and half dollars, amounts to $6.20. Find the number of each denomination. \/>-g = 8 \x+2y = ll 13. V?, .„ 16. 14. 15. 3/ + 4? = 10 }5;r-3 = 3j/ |- x +y = 3 J 3 x — y = — 2 2x + 3y = U 1? ' I*- 3.7 = 10 \2x + 5y = 12 j2a-8 = -b y-? jX =-l 1Sm 13^ + 4^ = 7 19 1 x Solving Equations with Two Unknowns 207 l*+8 = -4, on ff*+f>=12 ^ 20 < * s-' In the remaining examples of this exercise, use any method of elimination. 21. i^ + 4 . 25. J*=- 2 ^- 2;r+j/=16 ' 1 7 = * +14.5 >-j=10 26 * 12^-3^ = 7 ^=j/ + 2 * im \y + x =-lb 4^ = 3j/ + 3 \2x+2y = 24. { r / 28. 5y = 6x \x =y + 12 SUMMARY This chapter has taught three methods of solving a pair of equations which contain two unknowns : 1. Elimination of one of the unknowns by substitution ; 2. Elimination of one of the unknowns by comparison; 3. Elimination of one of the unknowns by addition or subtraction. REVIEW EXERCISE 94 1. If one tablet costs b dollars, what will x tablets cost? 2. If a books cost b dollars, what will one book cost ? c books ? 3. What is the perimeter of a rectangle whose width is a and whose length is b ? What is its area ? 4. What is the width of a rectangle whose perime- ter is/ and whose length is xl 208 Fundamentals of High School Mathematics 5. The area of a triangle is k. Its base is b. What is its altitude ? 6. The sum of two numbers is s. If one is d y what is the other ? Solve each of the following pairs of equations by any method of elimination : r — w 2 v = X r k ~ -. + 4=7 3,r+2; 7 f5* 6 4 T""8 = 8 8 [4^-^ = 10 10. A collection box contained 63 coins, nickels and quarters. How many of each kind were there if the total amount was $ 8.35 ? 11. Express algebraically that the weight, W, of the water in a tank varies as the volume, V, of the water. If 6 cu. ft. weigh 374.4 lb., how much will 11 cu. ft. weigh ? 12. In weighing with a spring balance scale, the principle is applied that the amount of stretch, s, of the spring varies as the weight, W, which is suspended to the scales. Express this more briefly. If a 20-pound weight produces a stretch of | inch, what weight will produce a stretch of 2 inches ? 13. What is the best method of eliminating one of the unknowns in an equation ? 14. The volume, V, of a sphere varies as the cube of its radius, r. Express this more briefly. If an orange with a radius of 2 inches is worth 10 cents, what could you afford to pay for an orange with a 4-inch radius ? Solving Equations with Two Unknowns 209 15. The table below shows how much money (to the nearest dollar) you would have at the end of a certain number of years if you saved 10 cents a day and deposited it in a bank which pays 3 % interest. At the end of (years) 1 2 3 5 8 10 14 17 20 total amt. saved is 37 75 H5 197 330 425 635 809 999 16. 17. Represent this graphically, measuring the time on the horizontal axis. Estimate the total savings at the end of -i yr. ; 6 yr. ; 25 yr. Is it ever possible to get from a graph informa- tion which could not be obtained from the table ? Illustrate. The distance from the base to the top of a hill, up a uniform incline of 40°, is 800 ft. What is the altitude of the top above the base ? CHAPTER XIII HOW TO FIND PRODUCTS AND FACTORS Section 99. Why you should be able to find products. Suppose you wanted to find the area of a rectangle whose X (SI 3X Area=6-X 2 Area= -3Xt5 Fig. 104 cm dimensions are 8;r + 5 and 2x. To do so, it would be necessary to multiply 3x+ 5 by 2x, or to find the product of these two algebraic expressions. One way to do this is to divide the rectangle into smaller rectangles, as indi- cated in Fig. 104. This gives two rectangles, the dimen- sions of one being 2x by 3x, and of the other 2x by 5. From what you already have learned about multiplication you can see that the areas of these are 6x 2 and 10. r, be- How to Find Products and Factors 211 cause Sx times 2x is 6x 2 and 5 times 2x is 10 x. Thus the area of the original rectangle is 6x 2 + 10 x. Similarly the area of the rectangle in Fig. 105 is what ? What would its area be if the dimensions were Qa + 4 and 5 a ? These illustrations are given to make clear the need of learning how to find products. Other illustrations might have been taken. For example, what is the cost of 15 b + 3 articles at 4 b cents each ? How much could you earn in 6 y + 4 days at Sy dollars per day ? A NEW WAY OF INDICATING MULTIPLICATION Section 100. As you progress through your mathe- matics, you will find that it is a language which tells more in fewer words or symbols than any other language. For example, instead of writing "find the product of 3x+5 and 2 x" it has been agreed to express this by means of the parentheses, ( ). Thus, 2.r(3;r+5) means " to find the product ofSx+o and 2x," or, "to multiply 3 x + 5 by 2x. It is important to note that there is no sign between the 2x and the expression in the parentheses. Similarly, to state algebraically the problem in the second illustra- tion, Fig. 105, you would write 4x(7x + 3), putting no sign between the 4:X and the parentheses. Thus, 5 b(3 b + 7) means to multiply each of the numbers in the parentheses by 5 b. ORAL EXERCISE 95 PRACTICE IN FINDING PRODUCTS In the following examples, multiply each term within the parentheses by the number which immediately precedes the parenthesis, or, remove parentheses. 212 Fundamentals of High School Mathematics Illustrative example. 3 *(5 x 1 + 7 x + 8) = 15 x 3 + 21 x 1 + 24 x. 1. 4 j'(3j' + 9) 10. 8 j/ (7 -8) 2. (i/;(i>/; + l) ii. 3r(r+3) 3. 7 £:(3 + 5 c) 12. lb{l-b) 4. f), r (.r-4) 13. 8(/; 2 -8£ + 12) 5. 9(2** + 7* -4) 14. 6(2*-3£ + <r) 6. 4a(a 2 +3a + 7) 15. ab{a + b + l) 7. 6j(3j> 2 - 5j/ + 2) 16. ^/ 2 (^r + j/ + ^) 8. 1(2^ + 3) 17. -1(4 -5 y) 9. 5/(6-/) 18. -3^r(6^-4) 19. What algebraic expression will represent the area of a rectangle whose length is 10 inches more than its width ? 20. What algebraic expression will represent the total daily earning of 4 men and 7 boys, if each man earns 1 2 per day more than each boy? 21. -6(2*- 7) 24. (16/ -7) 22. -(10-*) 25. (a + b) 23. _4j/ 2 (2j/-3) 26. -(b-c) In this exercise you have learned how parentheses are used to indicate that each of the terms in an expression must be multiplied by another number. Section 101. More difficult multiplication. Most prod- ucts which you will need to find are more difficult than those of the preceding exercise. For example : How many square feet of floor area in a dining room whose dimensions are 4^r + 3 ft. and 5x + 4 ft? To find the How to Find Products and Factors 213 T X 5 X -M--4- Area = 15 x Area = 20 X 2 Area=12 Area = 16 x -Sx+4 Fig. 106 product of these factors requires something you have not yet learned. You know how to multiply bx + 4 by 4 x or 4„r + 3 by 5x, but you have not learned how to multiply such expressions as 4^ + 3 by bx + 4. The drawing shows one way to do this ; namely, the geometrical method of dividing the entire area into smaller rectangles, the area of each of which you can find. This method gives four rectangles whose areas we can find. Thus, we get four rectangles whose areas are 20 ;r 2 , 16.r, 15 x y and 12, or, collecting terms, an entire area of 20 x 2 + 31 x + 12. Another method of finding the product of 4,r + 3 and 5x + 4 (which is generally written as (4„r + 3)(5x + 4)) makes no reference to rectangles. It is very much like the method of multiplication used in arithmetic. To illus- trate, this same example could be solved as follows : 2i4 Fundamentals of High School Mathematics First illustrative example. (4jc + 3)(5x + 4). 4jc + 3 5x-f-4 20x 2 + 15* + 16s -f 12 20jc 2 + 31jc + 12 The 20 ;r 2 + 15 .r is the result of multiplying 4.r + 3 by bx, and the 16^ + 12 is the result of multiplying \.x + 3 by 4. This latter method is much more generally used than the geometrical method. Let us take another illustration. Second illustrative example. (7 a + 3) (4 a -5). 7a + 3 4q-5 28a2 + 12 a - 35 a - 15 28 a3 - 23 a - 15 Note that 7 a + 3 was first multiplied by 4 #, giving 28tf 2 +12<*. Then 7 * + 3 was multiplied by -5, giving — 35 a — 15. How was the final product ob- tained ? Which of these two methods, do you think, should be learned ? How to Find Products and Factors 215 EXERCISE 96 PRACTICE IN FINDING PRODUCTS Illustrative example. (2a: 2 + 3x+4)(3x- 7). 2x 2 +3x +4 3s-7 6x 3 + 9x 2 + 12* -14s 2 - 21s -28 6x 3 - 5x 2 - 9x-28 1. (3^ + 2)(4^ + 5) 11. 2. (7£ + 5)(> + 6) 12. 3. ( 7 + 4)(5j/ + 3) 13. 4. (/-5)(2/+8) 14. 5. (8j/ + 5)(7j/ + 3) 15. 6. (6^ + 4)(4^ + 6) 16. 7. (4*-3)(4a-8) 17. 8. (*+3)<>r-9) 18. 9. (3^-4)(3^-9) 19. 10. (6/-7)(3/-9) 20. 21. 7 + 6)<j/ + 6) 3£ 2 +l)(2£2 + 5) 4^ 3 + 3)(3^ + 10) 5a+3)(5a -3) 7^ + 5)(7r — 5) # + 7)(<z — 7) 3^-2)(3^-2) x ?+2x+l)(x^-3) _T 2 + 6j/ + 9)(j/ + 3) ^ 3 + 5)(^ 3 -5) If you should multiply x + 7 by .r + 10, what would be the first term of your product ? What would be the last term ? 22. Can you tell, at a glance, the first and last terms of the product which you would obtain by multiplying 2x + 1 by 5 x + 4 ? A SHORTER METHOD OF MULTIPLYING Section 102. There is a much shorter method of finding products like those in Exercise 96. Mastery of this short 216 Fundamentals of High School Mathematics cut will not only save a great deal of time, but it will help you in the later work of this chapter. For an illustration, take the example (3£ + 5)(2£ + 7). You have no difficulty in seeing that the first term of the product is 6 b 2 {i.e. 3 /; x 2 b) and that the last term is + 35 {i.e. 5 x 7). So if there were some method by which you could tell the middle term, you could write the product at once, without using the longer method of placing one factor under the other and multiplying in the regular way. To make the new method clear, it is necessary to refer again to the regular way. Let us illustrate with the example : Illustrative example. (3ft + 5)(2ft+7). By the old method : 6 ft 2 + 10 ft + 21 ft + 35 6 ft 2 + 31 ft + 35 The arrows show the cross-muViplications or cross-products that make up the middle term. One " cross-product " is 2 ft times + 5, or + 10 ft, and the other cross-product is + 7 times + 3 ft, or + 21 ft. Combining the cross-products, we get the middle term, + 31 ft. By the shorter method we get + 21 ft (3ft+5)(2ft+~7) = 6 ft 2 + 31 ft + 35. + 10 ft The curved lines indicate the cross-multiplication, or cross-products, which must be combined to give the middle term, + 21 £ and +10£, giving +S16. Thus, in this How to Find Products and Factors 217 new method, it is assumed that you can tell, at a glance, the first and last terms of the product. Then you can get the middle term by finding the sum of the cross-prod- ucts. The curved lines are drawn to help you see the cross-products. Second illustrative example. (4& + 3) (7 &- 8). The Long Method The Short Method -32& (4 b +3) (7 b -8) = 28 b°— 11 b - 24 + 21b 24 28 fr 2 - 11 b - 24 It is important to recognize that the curved lines indicate cross-products in just the same way that the arrows in the long method refer to cross-products. The new method, which from now on we shall call the cross-product method, enables you to do mentally in much less time what was written down by the old method. To give you practice in this important method of finding the product of two factors the following exercise has been included. EXERCISE 97 Find the products of the following factors by the cross- product method : 1. (* + 2)(jr + 5) 6. (5£ + 3)(0 + l) 2. (2j/ + 4)(3j+5) 7. (9* + 2)(2;r+l) 3. (£+6)(2£+7) 8. (4^ + 3X7^+10) 4. (.r+4)(3.r + 5) 9. (5j + 3)(5j-3) 5. (/-8)(*-3) 10. + 9)0 + 9) 218 Fundamentals of High School Mathematics u. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. * + 2)(f— 5) 32. rt/> + <'>)(,?£ + 3) 33. .r 2 +3)(.r 2 + <;) 34. rt<V+8)(rt/;c-10) 35. ■2.V+ 3j')(2.r + 3j) 36. a + £)(a + £) 37. c + d){c + d) 38. f+s)(f+s) 39. .r 4- 5)(.r - 5) 40. ,r + 10)(,r-10) 41. J + 4)(j-4) 42. 3/ + 7)(3/-7) 43. 4rt4-3)(4rt + 3) 44. 7£ + 9j)(7£ + 9j/) 45. > + 3)(^ + 5) 46. x - 8)(x + 8) 47. c + 9)(c + 9) 48. ^-4)(7-4) 49. 3« + 2)(2a + 4) 50. 4 / -2)(3 7 + 5) 51. 2^ + 4)(5^--9) 52. y + 4)(rf+7) c- 9)^ + 9) 7 + 7)( 7 + 7) a - 2)(a - 2) 5a + l)(a-2) 2.r+l)(x-2) 4j + 3)(3j-9) /2 + 3)(7 2 + 5) /»-9)(/ + 2) r 2_3)( r 2_ 5) y + 3)(/ + 3) ^ 3 -ll)(^-ll) * 2 + 3)0 2 + 3) 4j+10)(5j-8) a^ + 2r)(rt^ + 3r) <6abc — d)(2abc + d) lx 2 t+y)(2x 2 t-5j/) 9.r 2 +3j)(9;r 2 + 3j/) 8«-4/)(8a-7^) /+ *)(/+*) How to Find Products and Factors 219 Area= X 2 + 12X+36 53. Illustrative example. What expression will repre- sent the area of a square if each side is x 4- 6 inches ? Solution : Evidently the area is (x + 6)(x + 6), or x°- + 12*4-36. This is usually written, however, not as (jc + 6)(x + 6) but as (x + 6) 2 . The exponent, 2. shows that x 4- 6 is used twice as a fac- tor. Thus, (3 x + 5) (3 x 4- 5) is usually written as (3 x 4- 5) 2 . In the same way. (3 x 4- 5) (3 x 4-5) (3x+5) would be written as (3x4-5)3. X+6 - Fig. 107 (0 54. + 5) 2 55. (3 £+5) 2 56. (2y + 7f 57. (4 a - 3) 2 58. (x+yf 59. {x-yf 66. 60. (a + *y> 61. (a- bf 62. (/4 ■sf 63. (/- sf 64. (2* + 3 y) 65. (4/; -2)2 (a +*)(« + d) Section 103. Further practice in translating from alge- braic symbols into word statements. You have had some practice in translating from algebraic statements into word statements. For example, you translated the alge- braic expression " a + b " into the word statement "the sum of two numbers," and the expression "xy" into "the product of two numbers." It is important to be able to translate into word statements some of the examples which you did in the last exercise. 220 Fundamentals of High School Mathematics EXERCISE 98 Write out the word statement which means the same thing as each of the following algebraic expressions : 1. a-b 7. (a + bf 2. a* 8. (a-bf 3. a* 9. ( a + b)(a-b) 4. (2j>f 10. r* + s s 5. c + d 11. (f+sf=f* + 2fs + s 2 6. a 2 + b 9 12. (f-sf=f*-2fs + s* Section 104. How to solve equations which involve prod- ucts. The solution of a great many equations depends upon your being able to find products like those you have just been finding. To illustrate, consider the problem : Illustrative example. The length of a rectangle is 6 inches more than, and the width is 2 inches less than, the sides of a square; the area of the rectangle exceeds the area of the square by 20 square inches. What are the dimensions of each ? Solution : Translating into algebra, we have the equation : (s+ 6)(s-2)=s 2 + 20. Multiplying, or removing parentheses, gives s 2 + 4 s — 12 = s 2 + 20. Subtracting s 2 from each side gives 4 s - 12 = 20. .-. s = 8. Thus, the sides of the rectangle are 14 and 8. Check this result. EXERCISE 99 PRACTICE IN SOLVING EQUATIONS WHICH INVOLVE PRODUCTS Solve and check each of the following equations : 1. ( jr + 6) 2 = ;r 3 + 96 2. (j' + 8)0' + 6)-y ! = 63 3. (2£ + 8)(£ + 4)=(£ + l)(2£ + 10) How to Find Products and Factors 221 4. One number is 5 larger than another ; the square of the larger exceeds the square of the smaller by 55. Find each number. 5. The length of a rectangle is 8 inches more than, and its width is 3 inches less than, the sides of a square ; the area of the rectangle ex- ceeds the area of the square by 26 square inches. Find the dimensions of the rectangle. 6. 0' + 5)0/-5) = (j/-6)( 7 + 2) 7. 2(x- 8)- 3(,r-4)=- ox 9. (^ + 3)2_(^ r _ 1)2 = 40 10. ( 7 + 5) 2 -(j + 4) 2 = -l 11. (x- 8f=(x- 12) 2 12. 2(/+3) 2 =2(/-8) 2 13. 3(x+6)(x + ±) = (Sx+l)(x+9) II. HOW TO FIND THE FACTORS OF AN ALGEBRAIC EXPRESSION A. FINDING THE COMMON FACTOR Section 105, Meaning of the word FACTOR. If you know that the area of a rectangle is 24 square inches, what might be its dimensions ? You readily see here that the dimensions might be 4 inches and 6 inches ; or 3 inches and 8 inches ; or 12 inches and 2 inches, because the product of 4 and 6, or of 3 and 8, or of 12 and 2, is in each case 24. This process of finding the members which, when multiplied together, give another is called FACTORING. The numbers you find are called the FACTORS. Thus, 4 and 6 are factors of 24 ; also 3 and 8, or 12 and 2. 222 Fundamentals of High School Mathematics Fig. 108 The same reasoning is used in algebra as in arithmetic. For example, suppose the area of a rectangle is 5^+35 square units. If the width is 5 units, what must the length be ? Similarly, if the area is 4:x 2 + 28 ;r and the width is 4;r, what is the length ? Section 106. What is a common factor ? Now let us take a more difficult illustration. In the previous two cases, you knew both the area and one dimension, or the product and one of its factors. But in most factoring prob- lems you do not know any of the factors. For example, what are the dimensions of the rectangle whose area is 7 x + 21, or, in other words, what are the factors of Ix -+-21? A study of the 7x and 21 shows that 7 is a factor of each, or is a COMMON FACTOR of both terms. What, then, must 7 be multiplied by to give 1 x + 21, or, what must be the length of this rectangle if its width is 7 ? Evidently, it must be x + 3. This prob- lem should be written 7^ + 21 = 7(> + 3). A second illustration should make clear what is meant by factoring in algebra. Find the factors of ax + ay + aw, or find the dimensions of a rectangle if its area is ax + ay + aw. How to Find Products and Factors 223 By observing each term, we see that a is a factor common to each. Dividing each term of the expression by a gives the other factor, x+y+w. Hence, ax + ay + aw = a(x+y + w), and a is one factor and x +y + w is the other one. These illustrations are intended to make clear how to factor an expression which contains a common factor. EXERCISE 100 Factor each of the following expressions : 1. 3^+12 15. 5x 2 —by 2 2. 5^-20 16. 4, a 2 + 8ab+4,P 3. ax + bx 17. X 2 + X 4. la -21 18. a 2 + 20 cP 5. llx + U 19. x 2 + 5x s 6. 6a + 9 20. a + ab + a 2 7. 5 + 15 a 21. x 2 -bx± 8. 9 + 6^ 2 22. 4a*-12aZ 9. 4^ 2 + 12 23. 5 x 2 y — 5 xy 2 10. Sa + 12b 24. 7b*-21 b 2 11. ab + ac + ah 25. 12ab +6# 12. 5 + 10 a + 15 a 2 26. x 2 + 2 xy 13. 2x 2 -±xy + 2y 2 27. 2- 20x 14. 2-8d 2 28. 64^ 2 -21^ 29. 6 ax 2 — 12 ax s Can you check the examples in this exercise? Check this one : 5* 2 -15;r 4 = 5;r 2 (l--3;r). j 24 Fundamentals of High School Mathematics B. THE (ROSS-PRODUCT METHOD OF FACTORING Section 107. In the previous section all the expressions which you factored had a common factor. But most ex- pressions which you will need to factor are much more difficult than those ; and they do not always contain a common factor. Illustrative example. Factor 2 x 2 + 5 x + 3, or find the di- mensions of a rectangle having this area. From what you learned about products, you can see that this expression was very likely made by multiplying two factors together. Also, you can see that the first terms of the two factors must be 2x and x. To help you to get the correct result, always write the blank form of the two parentheses first, thus: ( )( ). Then as you determine each term of each factor, you can write it in the appropriate place. Later you will doubtless be able to do all the work in your head and not have to write out the steps. Second, therefore, write the first terms in the blank form, thus : 2jc 2 +5jc+ 3=(2x )(x ). Thirdy you have to find the second terms of each factor. From your previous work in finding products, you know that the last term of the expression, + 3, was obtained by multiplying together the second terms of the two factors. Then the second terms must be such that their product is + 3. Obviously, they are 1 and 3 or 3 and 1. To tell whether the 3 or the 1 belongs in the first factor we have to try it, and test or check to see if the middle term will be correct (+ bx). Trying this out, we have : ex 2x 2 + 5jc + 3 =(2jc + l)(x + 3). x Haw to Find Products and Factors 225 Checking, — that is, multiplying the two factors together, — we see that this does not give the correct middle term, for + 6;r and + x are not ox. But, we might interchange the 1 and 3. Trying this, we get 2x 2x' 2 + bx + 3 = (2x + S)(x + 1). 3jc Multiplying these together shows that our factors are correct, for their product gives the original expression, 2. r 2 + ox + 3. Let us try another example : Second illustrative example. Factor 5 x- - 36 x + 7. First write the blank form thus : ( )( ). Next we can tell at once that the first terms of our factors are ox and x, giving 5x*-36x + 7 = (5x )( X ). Examining the last term of the expression, + 7, we see that the second terms of the required factors must be 1 and 7 or 7 and 1. Trying out the 1 and 7 gives 5x*-36x + 7= (5x+l)(x + 7). But the check shows that the sum of the cross-products is + 36 x, whereas it should be — 36 jr. This can be cor- rected by changing the sign of the second terms to — 1 and - 7, giving (5x-l)(x-7). The result of checking shows these to be the correct factors. 220 Fundamentals of High School Mathematics Section 108. The sum of the cross-products must equal the middle term. These two explanations have been given to show the importance of getting, as factors, expressions such that the sum of the cross-products will give the middle term of the expression to be factored. It is assumed that you can tell, at a glance, what the first terms might be, and what the second terms might be by looking at the first and last terms of the expressions which you want to factor. For example, in factoring 6;r 2 + 13.r+6, the first terms might be Sx and 2x } or 6x and x\ the second terms might be 3 and 2, 2 and 3, or 6 and 1. But since the product of the factors must give the original expression, we can tell by trying these various possible combinations that the factors are ±x (3* + 2)(2;r + 3), ~Yx No other arrangement of numbers will give the correct middle term, + 13 jr. From this explanation you should be able to factor the expressions in Exercise 101. Don't be discouraged if you have to try more than once before you succeed. Difficult tasks often require many trials. How to Find Products and Factors 227 EXERCISE 101 FACTORING BY THE CROSS-PRODUCT METHOD Check each example carefully. (The parentheses are written here to suggest to you how to begin.) 1. ** + 5*+6 = ( )( ) 2. 7 2 + 10j/-21 = ( )( ) 3. 2x 2 + 7x + 5 = (2x+?)(x+?) 4. c 2 + 6c + 9 = ( )( ) 5. zZ-8x+12=( )( ) 6. 5/ + I67 + 2 = (5. r+?)(j+?) 7. a 2 + 12 a + 36 = 20. 21 b 2 - b - 2 = 8. 5x* + 8x + 3 = 21. 5.r 2 -6^ + l 9. x*-Zx-U = 22. 3^2 + 4r + i 10. m 2 — m — 20 = 23. 2/ -j -28 11. 2 b 2 + 13 b + 15 = 24. 2 a 2 + 7 a + 3 12. 3^ 2 -13^-t-4 = 25. ^2_n r + 21 13. /*--5/-40 = 26. y _ i 0j/ + 20 14. IO7 2 + 13 jp - 3 = 27. ^ 2 + 6 a + 9 15. 4^ + 20^+25 = 28. y 2 -8y + 16 16. a 2 + # - 72 = 29. y + 10 17. y - 16 = 30. ^2 + ^ _ 30 18. £ 2 - 25 = 31. 3^r 2 + 8^ + 5 19. 15 c 2 _ 31 , + 14 = 32. 2^2_ 5;r + 3 Section 109. Importance of finding the prime factors. Any algebraic expression that cannot be factored is prime. For example, 3 x + 5 is prime, because there are no integral expressions which can be multiplied together to produce it. 228 Fundamentals of High- School Mathematics But 9jt+6 is not prime because it can be obtained by multiplying 3 and x + 2. It is important that you should always find prime factors. To illustrate : First illustrative example. Factor 3 b 2 - 21 b + 36. By inspection, we see that 3 is a common factor. Removing it, we have S(b 2 - 7 6+ 12). Now. unless we remember that prime factors should be found, we are likely to leave the example in this incomplete form. The b 1 — 7 b + 12 can be factored further, however, giving (*-3)(*-4). Thus, the original example should be factored as follows : 3 b 2 - 21 b + 36 = 3(62 - 7 b + 12) = 3(6-- 4) (b -3). Second illustrative example. Another illustration will make clear the importance of finding prime factors. Suppose we wish to factor 2 x 2 — 50. As in the previous example, we always first look for a common factor. This gives 2(x*-25). Now, again, we are apt to leave the example in this incomplete form, not remembering to see if we can further factor x 2 — 25. We see, however, that we can. The complete solution is : 2x 2 -50 = 2(x 2 -25) = 2(x + 5)(x-5). These explanations are given to help you keep in mind that in all factoring work there are two absolutely essen- tial steps ; namely, 1. LOOK FOR A COMMON FACTOR. 2. FIND PRIME FACTORS; I.E. FACTOR COMPLETELY. How to Find Products and Factors 229 1. 2^ 2 + 14^4-24 2. 5j/ 2 -45 3. sfl - st- 20 s 4. 7^-14^-105 5. 3*3 + 12* + 45 6. * 2 -6*+9 7. 6/ 2 -15 / 3 8. 2 - 128 / 2 9. aP-a&- 72 a 10. 6* 2 + 13* + 6 11. 3 a 2 + a - 2 12. 2 ^ - 5 * + 3 13. 7 £ 2 - 17 £-12 14. <f _ 12 £ - 28 15. 2* 2 -14* + 24 16. ^ 4 — 6 j/ 2 — 16 17. 49^+70^+25 18. 5a 2 - 80 19. x z — X 1 20. 6* 2 -18* 3 21. 3y _ 3 y _ 36 22. 12* 2 + 37*-10 23. 10* + 25+* 2 24. j 2 + 10 25. * 2 + 36 EXERCISE 102 PRACTICE IN FACTORING COMPLETELY 26. 2 a 2 -8 27. 3£ 3 + 27 28. 3^-12^-180 29. 2 ^ + 10 x -168 30. ;r 2 -.r-110 31. 2j/ 2 -j/-1 32. 6 a 2 -4a -2 33. 3x 2 + ±x + l 34. 20.r 2 + 70.i- + 60 35. 2£ 2 -£-3 36. 2rt 2 + 18 37. 5 y — 15 j 2 38. a 2 -b 2 39. 3 J 2_ J _ 10 40. 2*2 + 3^_9 41. 6j/ 2 +7— 15 42. 8* 2 -2 43. 18 fl - 50 44. 9* 2 + 17*-2 45. 6^_^_ 12 46. 7 y _ 9 y _ 10 47. 2^-36* + 64 48. j/ 2 — 3 j/ — 4 49. / 2 -16 50. / + 16 230 Fundamentals of High School Mathematics SUMMARY This chapter has taught the following methods : 1. A short method of multiplying which we call the " cross-product " method. 2. How to find the factors of an algebraic expres- sion. REVIEW EXERCISE 103 1. What is the area of a square formed by adding 4 ft. to the sides of a square x ft. long ? 2. What does (x— 4)(;r -f 6) represent, if x repre- sents the side of a square ? 3. A rectangular field by rods long has a perimeter of 24 j rods. What expression will represent the area of the field in square rods ? in acres ? 4. If the quotient is represented by q, the divisor by d y and the remainder by r, what will repre- sent the dividend ? 5. If a park is w rods wide and 1 rod long, how many miles would you walk in going around it n times ? 6. How do you divide a product of several factors by a number ? For example, in dividing 12-3-6 by 2, would you divide each factor by 2 ? 7. How do you multiply a product of several fac- tors by a number ? Give an illustration. 8. The product of four factors is 60. Three of them are 2, 3, and 5. Find the fourth factor. 9. How much do you increase the area of a square whose side is x, if you increase its side 4 units ? How to Find Products and Factors 231 10. Translate into words : (/+ s) 2 =/ 2 + 2fs + s 2 . 11. Solve for x y explaining each step : -4^+6 = 2^-18. 12. In what way is factoring like division ? How is it like multiplication ? 13. If an automobile uses 8| gallons of gas in going 120 miles, how many gallons will it use in going 250 miles ? 14. Solve 1 J 6 y -j=T + 4 y y I5;r + 8j/=l. 15. Make up five examples for the class to factor, and then give them ^o the class to work. 16. How many terms do you get when you square the sum of two members, e.g. (2x+3yf? when you square the difference of two numbers, e.g. (4 * - 3 bf ? 17. Evaluate {a + bf if a = — 3 and b = + l. 18. Does (a + bf = a 2 + b 2 ? Show by using 4 for a and 5 for b. 19. If — 1 = ^2, what is the value of V x when V 2 V 2 Pi = 40,^ = 8, and/ 2 = 12? 20. A tree stands on a bluff on the opposite side of a river from the observer. Its foot is at an elevation of 45° and its top at 60°. Which has the greater height, the bluff or the tree ? What measurement would you have to make to find the height of the tree ? the width of the river ? 21. Translate : a 2 - b 2 = (a + b)(a - b). CHAPTER XIV HOW TO SOLVE EQUATIONS OF THE SECOND DEGREE Section 110. What are quadratic equations ? In all the previous chapters you have solved equations of the first degree ; that is, equations in which the unknown (or unknowns} did not have exponents greater than 1. This chapter will show how to solve equations of the second de- gree, equations in which the unknown occurs to the second power. To illustrate, you will learn how to solve equations such as x 2 + 8.r=20. The fact that the unknown, x, in this equation occurs in the second power or second degree (as, x 2 ) leads us to i speak of the equation as a second-degree, or quadratic, EQUATION. Three ways of solving second-degree or quadratic equa- tions will be explained. These, in the order in which we shall discuss them, are : I. Solution by graphical representation. II. Solution by factoring. III. Solution by completing the square. Before we take up the first method, it will be necessary to study how to represent graphically algebraic expressions of the second degree which are not equations. A. HOW TO REPRESENT GRAPHICALLY AN EXPRESSION OF 'I'Hi; SECOND DEGREE IN ONE UNKNOWN Section 111. The value of an expression depends upon what value is assigned to x. Let us consider the second degree or quadratic expression 2X2 How to Solve Equations of the Second Degree 233 Has this expression a definite numerical value ? What is it? 10? 18? If not, what? Evidently, we cannot tell what the value of the expression is unless we know what the value of x is. That is, the numerical value of the expres- sion depends upon what value is assigned to x. In other words, as x changes, the value of the expression x 2 — 8x + 12 changes. Thus, we are dealing here with two vari- able quantities ; x itself is one variable, and the value of the expression jc 2 — 8 x + 12 is the other one. Section 112. (1) Tabulating values of the two variables. Thus, to represent an algebraic expression of the second degree graphically we first have to determine what numeri- cal value the expression has for various assigned values of x. That is, we will let x be 0, say, and we find by evalu- ating x* — 8x + 12 when x = 0, that the expression is 12. In the same way, when x = 1, the expression is 5 ; when x = 2, x*-8x+12 is 0, etc. We have already learned that to represent the way in which two related quantities change together, it is best to tabulate first the values of the two variables. For the present example we get the following table : Table 15 If xis 1 2 3 4 5 6 7 -1 -2 then X-8X+12 is 12 5 -3 -4 -3 5 21 32 etc. Section 113. (2) Plotting the graph of the two variables. With the table of values of the two variables [(1) the un- known, (2) the expression which contains it] once made, we can plot the points representing these pairs of values. Let us agree for convenience to plot the unknown, say x. 234 Fundamentals of High School Mathematics is ,2 on the horizontal axis and the expression (in this case, x- — &x+ 12), on the vertical axis. Reading the pairs of values 0, 12; 1, 5 ; 2, ; etc., from the table, we obtain as the graphic representation of x 2 — §x+12 the curve of Fig. 109. Note that from the graph we can find the value of x 1 — Sx + 12 for any value of x represented on the .f-axis. For example, if x is Y 2 J, then ^ 2 -8^+12 3| ; if x is 0, then 8;r+12is 12, etc. Note also from the curve that x 1 — 8 x + 12 is zero for two particu- lar values of x\ namely, x= + 2 and x= +6, i.e. tJie curve cuts the x- axis at + 2 and + 6. The graph shown in Fig. 109 is called a graph of an algebraic expression of the sec- ond degree, or a quad- ratic expression. Section 114. Graphs of second-degree equations are always curved lines. Turn back to Figs. 14, 16, 17, etc. What difference do you notice between each of the graphs in these figures and the one in Fig. 109 ? If you will turn to Fig. 110, you will check your conclusion. One new fact of importance is, therefore, that the graphs of expressions of the first degree are always straight lines, whereas graphs of expressions of the second degree are curved lilies. i i , 1 \ , 1 < *J i 1 r • f y A i 4d \ / / \ » r / i H If V ! ( ~\ I V j \ 1 s i Fig. 109 How to Solve Equations of the Second Degree 235 +"( ) 1 \ / I / \ / \ J 1C -5 r ) b 10 15 I \ | 1 i "lC ) J \ i 1 1 1 h s \ 1 j -2 c ) ! t \ 1 A V 'V \ L / ^ / ^ l -'< ) / *s f \ / s -/I 3 -X Fig. 110 Graphs of such quadratic expressions are U-shaped and are commonly called parabolas. Figure 110 is a graphical representation of the quadratic expression x l — 2x ■— 35. By referring to it, you will be able to answer the following questions ; EXERCISE 104 1. How many different values may the expression have ? 2. What is the value of x when the value of the expression is zero ? 3. For what value or values of x is the expression equal to — 10 ? 236 Fundamentals of High School Mathematics 4. What is the lowest or least value of the ex- pression ? 5. I low were the points which fix the position of this curve located ? Are two points enough to determine the graph ? 6. What is measured along the horizontal axis ? the vertical axis ? 7. What variables are represented in this graph ? ■ EXERCISE 105 PRACTICE IN CONSTRUCTING GRAPHS OF QUADRATIC EXPRESSIONS In each case, determine from the graph what values of x % if any, will make the expression equal to zero : 1. ^-6*+ 8 5. x^-lx-Xb 2. ^2_4 r + 3 6. ^ 2 -9 3. ^2 + 7^+10 7. x 2 -8x + 16 4. x 2 + x _ 12 a x * + 4:x+2 Now that we have seen how to graph a quadratic EXPRESSION, we are ready to consider the first method of solving quadratic equations : B. HOW TO SOLVE QUADRATIC EQUATIONS I. GRAPHICAL SOLUTION Section 115. To solve a quadratic v equation is to find values of x which will make the expression equal to zero. In the previous section we graphed quadratic EXPRESSIONS. It is important to note that they were not quadratic EQUATIONS. Thus, ^r 2 -8^+12 is a quadratic expression, — that is, an algebraic expression of the second degree in one unknown, — but it is not an eqita- How to Solve Equations of the Second Degree 237 Hon. If, however, we should write x 2 — 8x + 12 = 0, then we have a quadratic equation, — an equation of the second degree. To solve this equation is to find the value or values of x which will make x 2 — 8;r+12 equal to zero, because such values will make one member of the equation equal to the other ; i.e. = 0. Graphically, this can be done by plotting the values of the expression x 2 — 8x+ 12, as we did in the previous section. Thus, to graph the quadratic equation x 2 — 8.r + 12 = 0, we graph the ex- pression x 2 — 8 x + 12, and note from the graph what values of x will make the expression equal to zero. Thus, from the graph we see that x 2 — 8x + 12 is zero when x = 2 or when x = 6. Thus, we see that x can have two values in the quadratic equation ^2 -8*+ 12=0. Checking, we find that 2 and 6 each satisfies the equation. FlG m This suggests the following method for solving a quadratic equation graphi- cally : 1. Graph the quadratic expression which forms one member of the equation. (The other member should be zero,) 1 1 , 1 1 , ' r J 1 j ' ■ i ! ' X L_ <b V / / \ / < V T^ h <i \ { "1 i V J \ ' , 1 !±S 238 Fundamentals of High School Mathematics '1. From the graph find for what values of x the expression is zero. In other words, find at ivJiat value of x tlie graph cuts the x-axis. These values are the values of x which will satisfy the equation. 3. Check your result by substituting these values of the unknown in the original equation to see if they do satisfy it. EXERCISE 106 PRACTICE IN SOLVING QUADRATIC EQUATIONS GRAPHICALLY Solve the following quadratic equations by drawing graphs of the expressions : 1. x 2 - 5x - 14 = = 6. 2/+ 5/4-3 = 2. A -2 + 3^=40 7. x* + 8x + 16 = 3. f - y = 20 8. f + By + 1 = 4. 2x = 48 - x 2 9. O- 2) 2 + 6;tr = 12 5. x 2 -6x+9 = 10. x 2 + 4 = In these examples, did you find any graph that did not cut the .r-axis in two places ? What conclusion would you draw if the graph just touched the ;r-axis ? if it did not even touch it ? What seem to be the disadvantages of the graphical method of solving quadratic equations/? the advantages ? How to Solve Equations of the Second Degree 239 II. HOW TO SOLVE QUADRATIC EQUATIONS BY FACTORING THE EXPRESSION Section 116. By making use of a principle that you al- ready know, — namely, that the product of any number of factors is zero if one of them is zero, — we have a very easy method of solving quadratic equations. Recall that 4.5.0 = 0, or that 6 • • 26 = or that a ■ b • c = if either a, or b, or c is 0. Why ? Because any number multiplied by is 0. Under what conditions is the prod- uct, xyz, zero ? Evidently, if eitlier x, y, or z is zero ; that is, if one of the factors is 0. In the same way, the expres- sion (x — 4)(;r+ 6), which is the product of two factors, could be if x— 4 were 0, or if x + 6 were 0. Now let us apply this to the solution of a quadratic equation. Illustrative example. The square of a certain number, in- creased by twice the number, gives as a result 48. What is the number ? Translating into algebraic language, we have the quadratic equation x 2 + 2* = 48. (1) In order to solve the equation we want to get a product equal to zero. Hence the equation x 2 -f 2 x = 48 should read, x 2 + 2* -48 = 0. (2) Factoring the left member to form a product, (x + 8)(x _ 6)=0 . (3) Note here that we have a product equal to 0. But, in order that a product can be 0, one of the factors must be 0. Therefore, if x + 8 = 0, x must equal - 8, or if x — 6 is to be 0, x must be 6. Summarizing the solution, we have the following steps : x 2 + 2jc=48. (1) Making one side 0, x 2 + 2 x — 48 = 0. (2) Forming a product, (x + 8)(jc - 6)= 0. (3) Making each factor 0, x + 8 = 0, or x = — 8. (4) x-6 = 0, orx=6. (5) Checking, by substituting 6 and — 8 in equation (1), 64 - 16 = 48, or 36 + 12 = 48. 240 Fundamentals of High School Mathematics EXERCISE 107 Solve the following quadratic equations by factoring. Check each one. 10. y + ^ = e 2 11. a 2 + a = 20 13. r 2 +2^=0 14. ^ 3 -2.r 2 = ^r 15. 12 = ^ 2 + ^ 16. The altitude of a triangle exceeds its base by 4 inches ; the area is 96 square inches. Find the base and altitude. 17. Find two consecutive even numbers whose product is 120. 18. The sum of two numbers is 10 ; the sum of their squares is 52. Find each number. 1. *a-5;r + 6 = 2. ^2_ J/ = 20 3. * 2 + 9a = 22 4. b 2 - 36 = 5. 2* 2 -5;r + 3 = 6. e 2 - 8e= -16 7. VI 1 =7/1 + 2 8. fl 4- / = 56 9. i^2 +r = 12 Which of the two methods of solving a quadratic equa- tion that we have considered thus far do you think is better ? State the reasons for your choice. Could you solve any quadratic equation by the Factoring Method ? Solve x 1 + 3 x = 12 by this method. Before we can consider the third method of solving quadratic equations, — namely, solution by " completing the square," — it will be necessary to learn how to find the square root of algebraic expressions. How to Solve Equations of the Second Degree 241 SQUARE ROOT Section 117. The need for square root. In arithmetic you learned how to find a side of a square when its area was known. For example, if you knew that the area of a square was 81 square inches, you learned to find one side of the square by finding one of the two equal factors of 81, or, in other words, by finding the square root of 83 . Just now we cannot learn the best method of solving quad- ratic equations until we learn how to find the square root of algebraic expressions and of arithmetic numbers. Section 118. What is square root? The square root of a number is one of the two equal factors of that number. Let us illustrate with the number 100. We might factor it in several ways, as follows : 4 x 25 = 100 5 x 20 = 100 10 x 10 = 100 8 x 12.5 = 100, etc. Remembering that the square root of a number is one of its two equal factors, we see that neither 4 nor 25 is the square root of 100, because they are not equal factors. Evidently, 10 must be the square root. In the same way, — 10 is also a square root of 100 because it is one of two equal factors of 100. In the same way the cube root of a number is one of the tliree equal factors of the number. Thus, — 2 is the cube root of - 8 because (- 2)(- 2)(- 2) = - 8. Section 119. How to indicate the root of a number. It has been agreed to indicate the root of a number by the symbol V~ , called a radical sign. To designate what 242 Fundamentals of High School Mathematics root is meant, a small number called an index, is placed in the radical sign. Thus, VlO means the fourth root of 1(), i.e. 2, because 2 • 2 • - • - = 1(> ; V27 means the cube root of -7, i.e. 3, because 3 • 3 . 3 = 27 ; and V25 means the square root of 25. It is customary, however, to omit the index when square root is meant. Thus, V25, without an index, is always understood to mean the square root of 25. EXERCISE 108 FIND, BY TRIAL, THE ROOTS,. WHICH ARE INDICATED, OF THE FOLLOWING EXPRESSIONS i. V9 io. V/ f 2. V^« ,, J36.r 2 3. vsra ' , »•>* 4. *§ 12 ^^_ v 13. V144^ 10 5. v 27 x* / <AA 9 . , i4. V4oo.f 2 y 6 ' ^ la 15. ^125^? 7. V64* 8 16 </i6^i 8. VlOOy 5 i7. -v / 243y° 9. V] 18. a/25G x 12 Section 120. How to find the square root of algebraic expressions. We have seen that {a 4- bf or {a + b)(a + /;) = a 1 + 2 ab 4- b 2 . From this iV is evident that the square root of a 2 + 2 ab + b 2 must be a + b, or V# 2 + 2 ab + £ 2 = (<? + b). In the same way Vx 2 + 10 x + 25 is x + 5, be- cause (/r + 5)(.# + 5) gives .r 2 + 1 x + 25. From these illustrations we see that it is possible to extract the square How to Solve Equations of the Second Degree 243 root of an algebraic expression if we can show that it can be obtained by squaring some other expression ; that is, if we can show that it is the product of two equal factors. EXERCISE 109 Find the square root of the following expressions, where it is possible to do so. Check each. 1. x 2 + 2xy + y 2 8. y 2 + 6y + 20 2. ;z 2 + 6rt + 9 9. 25^ 2 + 40^ + 16 3. p _ 4 b + 4 10. .r 2 + 16 4. / 2_ 10/+ 25 11. y 2 ~ 49 5. 4<z 2 + 12tf + 9 12. 12^ + 36+^ 2 6. 1Cd+x 2 + 8x 13. t 2 + u 2 + 2 /» 7. 1 + 21*+ 100 * 2 14. r¥ - 6 n- 2 + 9 If any of the expressions above are not perfect squares, make the necessary changes to transform them into perfect squares. HOW TO FIND THE SQUARE ROOT OF ARITHMETICAL NUMBERS Section 121. We have learned that the square root of a number is one of its two equal factors. Hence, to find the square root, we need to find one of its two equal factors. If a number is divided by its square root, the quotient will be the same as the divisor. Thus, if 36 is divided by its square root, 6, the quotient will be the same as the di- visor, 6. But if 36 is divided by a number which is smaller than its square root, the quotient will be larger than its square root. Thus, if 36 is divided by 4, the quotient is 9. The square root is somewhere between the divisor, 4, and the quotient, 9. If the divisor and quotient are the same, 244 Fundamentals of High School Mat Hematics either one is the square root of the number, but if they are not the same, then the square root is some number between them. The next exercise will illustrate this trial method of finding square root. Find the square root of each of the following numbers by the trial method : First illustrative example. Find the square root of 55. Since this number is between the two perfect squares, 49 and 64, its square root will be between 7 and 8 ; let us try 7.5 and use it as a divisor (one factor) to find the quotient (the other factor). 7.333 7.50)55.0 52 5 2 50 2 25 250 225 250 This shows that the square root of 55 is between 7.500 and 7.333. Let us try a number halfway between them, say 7.416. 7.4164 7.4160)55.000 51 912 3 0880 2 9664 12160 7416 47440 44496 29540 29664 The square root, then, of 55 is 7.4162 This can be checked by multiplication or by division. How to Solve Equations of the Second Degree 245 Second illustrative example. Find the square root of 12£. Since the number is between the two squares, 9 and 16, its square root will be between 3 and 4; let us try 3.5 and use it as a divisor. 3.57 3.5)12.5 10 5 2 00 1 75 250 245 5 This shows that the square root of 12.5 is between 3.50 and 3.57. A value closer than either of these is the number halfway be- tween them, say 3.535. 3.536 3.535)12.5 10 605 1 8950 1 7675 12750 10605 21450 21210 This shows that the square root is between the two factors 3.535 and 3.536. Using the number halfway between them, 3.5355, we have the square root correct to 4 decimals. EXERCISE 110 By using this method, find the square root of each of the following numbers : 1. 18 4. 500 7. 965 10. 14.75 13. 3 2. 52 5. 16.80 8. 3820 11. 2025 14. 2 3. 200 6. 150 9. .2640 12. 8 15. 10 246 Fundamentals of High School Mathematics Section 122. The old method of finding square root. 1 Many pupils have learned in arithmetic another method of finding the square root of numbers. It is illustrated in the following example : Illustrative example. Find the square root of 200. Note the following steps : 200.0000)14.14 (1) The number is separated into periods L of two figures each, counting from the decimal point. 24)100 (2) You find the greatest square in the gg left-hand period, and write its square 9Q 1 /LAO root for the first figure of the root. ^ c ' (3) Subtract this square from the left- 281 hand period, and with the remainder 11900 place the next period for a new divi- 2824)11296 dend. (This is 100 in the example.) chyL (4) Double the part of the root already found (2x1 = 2) for your trial divisor. Divide the dividend, exclusive of the right-hand figure (10) by the trial divisor, 2. Write the quotient obtained, 4, as the next figure of the root and the divisor. Multiply the complete divisor, 24, by the last term of the root, 4. Subtract the product, 96, from the dividend, 100. To the remainder, 4, annex the next period, 00 for a new dividend. Repeat this process until all periods are used, or until any required degree of accuracy is obtained. EXERCISE 111 Solve the examples of Exercise 110 by this method. 1 It is believed that this traditional method is much more difficult to rationalize for the pupil than the so-called trial or estimate method. For those teachers who insist upon its use, the trial method may be omitted. In the interest of experimentation, however, the authors hope the proposed method will be fairly tested out. How to Solve Equations of the Second Degree 247 III. QUADRATIC EQUATIONS SOLVED BY THE MOST GENERAL METHOD: COMPLETING THE SQUARE Section 123. The third method of solving quadratic equa- tions. Only a few easy quadratics can be solved by the second method which we studied, i.e. only those which can be factored. Furthermore, the first method shows that graphical solutions are too slow and often give only ap- proximate results. Consequently we need a more general method — one that is applicable to all quadratics, and one that gives accui'ate results. Let us illustrate such a general method of solving quadratic equations. First illustrative example. The area of a rectangle, which is 6 inches longer than it is wide, is 55 square inches. Find its dimensions. Translating into algebraic form, we have x 2 + 6x = 55. (1) Adding 9 to each side to make the left side a perfect square, x 2 + 6 x + 9 = 64. (2) Extracting the square root of each side, x + 3 = + 8 or - 8. (3) Using the +8, x = 5. (4) Using the -8, x=-li. (5) Checking the result : etc. Section 124. We must make the left side a perfect square. It is important to see why 9 was added to each side of equation (1). Why not add 10 or 20 or any other number, to each side ? Because tJie left side would not be a perfect square if any other number were added. Success with the method of completing the square depends upon knowing what number to add to each side to make the left side a perfect square. Note that the left side must be a perfect square, because you cannot extract the square root of a?i algebraic expression zvliich is not a perfect squa7'e. 24^ Fundamentals of High School Mathematics Second illustrative example. Find a number such that its square decreased by 3 times itself shall be 10. Translating into algebraic form gives : x 2 -3;t = 10. (1) Adding $ to each side in order to make the left side a perfect square, X2_ 3x + } = 10 + £ = ^. (2) Extracting the square root of each side, x - f = | or - f. (3) Using + }, x = Af- or 5. (4) Using -I, x = - I or - 2. (5) Check : etc. .*. The unknown number is either -f- 5 or — 2. Here again, as in the first illustrative example, the most important and most difficult step is completing the sqziare, i.e. to know what to add to x* — 3 x to make it a perfect square. We need to know why |, rather than some other number, was added to each side. Remember that the algebraic expression, the left side of the equation, must be a perfect square, for otherwise we could not extract the square root of it. But we can extract the square root of the right side (because it is an arithmetical number) even if it is not a perfect square. Section 125. How to complete the square of any quadratic expression : add the square of one half the coefficient of x. Because we use the method so frequently, it will be worth while to learn the general method of completing the square. p If we square x + ^ , we get the expression that is, ( x + £) = x(2, +P x + ^t' How to Solve Equations of the Second Degree 249 Note here that the last term of the expression is ^-- It P is the square of —' But p represents the coefficient of x in the expression which we wish to change into a perfect square. Thus, if we had the expression x 1 +$x and desired to change it into a perfect square, i.e. to com- plete the square, we should have to add — ; that is, add the square of one half the coefficient of x. To complete the square in the expression x 2 + 6x (see first illustrative example), we should add the square of one half the coefficient of x ; that is, the square of J of 6, or 9. In the same way, to complete the square in the expression x 2 — 3 x, we must add the square of one half the coefficient of x\ that is, the square of \ of 3 or (f) 2 , which is f. EXERCISE 112 Solve each of the following quadratic equations by the method of completing the square : 1. x 2 + 6x = ±Q 6. j 2 + 7j=8- 2. * 2 -8;r=84 7. x 2 + x=56 3. f + 2y=lb 8. ^ 2 +6^ = ll 4. ^2_ 10 ^ = _ 1(3 9 p2_Ap = ^ 5. /2 + 3/ = 10. 10. 10.r + ^ 2 =-9 11. A rectangular field is 2 rods longer than it is wide, and it contains 6 acres. Find the length of the sides. 12. Find two consecutive even numbers whose prod- uct is 80. 250 Fundamentals of High School Mathematics 13. Find the value of x in the equation Illustrative example. 3x 2 -13x + 4 = 0. Solution : The equation must be in the form jc2 + />jc = n. In other words, the coefficient of x 2 must be 1. Dividing each side of (1) by 3 gives Adding (\ 3 ) 2 to each side gives v2_I3y J. 1_6 9 — 133. — 4. — 121 * 3 x * 3^ — 3¥ 3 — 35 ' Extracting the square root of each side gives *-¥ = -¥ or -y. Using + V, x = ¥ + ¥ = ¥ = 4. Using --V-, x =¥-¥ = f = i Substituting J for x in equation (1), to check gives 3.^-13.^ + 4 = 0. 1 - ¥ + ¥ = o. (i) (2) (3) (4) The pupil should check for x ■ 14. 2x 2 + 10 x=12 15. 3a 2 + 6a = 45 16. - + - = 9 2 4 Hint: Get rid of fractions. 17. y-40 = 8y 18. 20. 2;r 2 3 x 3 — ;r= 3 19. x — 47 2 * 10 ^,§£^81 2 5 2 21. 3j 2 + 5j = 22 22. 5£ 2 +lG£+3: 23. The difference of two numbers is 4, and the sum of their squares is 210. Find the number. 24. A farmer has a square wheatfield containing 10 acres. In harvesting the wheat, he cuts a strip of uniform width around the field. How How to Solve Equations of the Second Degree 251 wide a strip must be cut in order to have the wheat half cut ? 25. Divide 20 into two parts whose product is 96. 26. The sum of two numbers is 20, and the sum of their squares is 208. Find the numbers. 27. I went to the grocery for oranges. The clerk said they had advanced 10 cents per dozen. I got J dozen fewer oranges for a dollar. What was the original price per dozen? 28. A piece of tin in the form of a square is taken to make an open-top box. The box is made by cutting out a 3-inch square from each corner of the piece of tin and folding up the sides. Find the length of the side of the original piece of tin if the box contains 243 cubic inches. 29. A rectangular park 56 rods long and 16 rods wide is surrounded by a boulevard of uniform width. Find the width of this street if it con- tains 4 acres. 30. The members of a high-school class agreed to pay $8 for a sleigh ride. As 4 were obliged to be absent, the cost for each of the rest was 10 cents greater than it otherwise would have been. How many intended to go on the sleigh ride ? 31. Solve by all three methods the following quad- ratic equation : 2^ 2 + 5^ = 18. Fig. 112 252 Fundamentals of High School Mathematics REVIEW EXERCISE 113 1. The length of a 10-acre field is 4 times its width. What are its dimensions ? 2. Does x = — I satisfy the equation 4 x 1 + 11 x= -6? 3. The space passed over by a body falling t sec- onds is expressed by the formula 5 = 16/ 2 , where 5 is the number of feet the body falls. Construct a graph for this formula, using for t the values 0, J, 1, 11 2, 2J, and 3. Plot the values of / along the horizontal axis. 4. Does V9 + Vl6 = V25? Does the sum of the square roots of two numbers equal the square root of the sum of the numbers ? Does Va + V6 = Va + 6? 5. Is the square of a number always larger than the number ? Illustrate. 6. Evaluate -\/s(s — a)(s — 6)(^ — c) if ^ = 20, a = 8, b = 14, and <: = 18. 7. What is the square root of — 25 ? of — x 2 ? 8. Does (x -f j/) 2 = .r 2 + jj/ 2 ? Test by using 4 for x and 3 forjj/. 9. If you wanted to divide a product of several factors, such as 6 • 8 • 10 • 12, by some number such as 2, would you divide each of the factors by 2? 10. Complete the statement: To divide a product by a number, divide of the factors by that number. How to Solve Equations of the Second Degree 253 11. If a square piece of tin 10 inches on each side sells for 60 cents, what should a 15-inch square piece of the same thickness sell for ? 12. A girl went to the bakery for pies. If she could buy pies 5 inches in diameter for 15 cents each, or 10 inches in diameter (the same thick- ness) for 35 cents, which would be cheaper for her to buy ? SUMMARY This chapter has taught : 1. The meaning of a quadratic equation. 2. How to solve quadratic equations by three methods : (a) By graphical representation. (6) By factoring. (c) By completing the square. 3. Square root. CHAPTER XV FURTHER USE OF THE RIGHT TRIANGLE: HOW TO SOLVE QUADRATIC EQUATIONS WHICH CONTAIN TWO UNKNOWNS Section 126. Previous use of the right triangle. We have already seen how right triangles can be used to find unknown distances. If we knew one side, and one acute angle, we were able to find any other side. Now we come to another method of dealing with right triangles ; namely, when two sides are known, but when no acute angle is known. This method will be illustrated by the following problem : What is the longest straight line you can draw upon a rec- tangular blackboard 28 in. wide and 36 in. long ? Evidently the longest straight line is the diagonal of the blackboard, or the hypotenuse of the right tri- angle, Fig. 113. Thus, we need to know how to find the hypotenuse of a right triangle when the other two sides are known. Fig. 113 This leads to the following : THE PYTHAGOREAN THEOREM Section 127. This important relation between the sides of a right triangle was discovered by the celebrated Greek mathematician Pythagoras, after whom it has been named. This theorem or law states that the square of the hypote- nuse of a right triangle is equal to the sum of the squares of the other two sides, or, in the above problem, that h* = 36 2 + 28 2 . 254 The Pythagorean Theorem 255 A Fig. m This relation between the sides of a right triangle can be seen from Fig. 114. The base, AB, and altitude, AC, of a right triangle are drawn so that they contain a common unit an integral number of times. AB con- tains the common unit 1 times and AC contains it 3 times. Then by actual measurement BC will con- tain the same unit 5 times. By constructing squares on the sides of the triangle, you can see by counting that the sum of the squares on AB and AC is equal to the square on BC. To test this further, the pupil should construct a right triangle with the base 12 units and the altitude 5 units. Then actually measure the hypotenuse, and note whether the square on the hypotenuse is equal to the sunt of the squares of the other two sides. Now we are ready to go back to the problem of finding the longest line that can be drawn upon the blackboard. By making use of the truth which was just studied we have : h 2 = 28 2 + 36 2 or h 2 = 781 + 1296 or h 2 = 2080 or y£=V2080 = 45.66 in. This relation between the sides of a right triangle is more widely used by engineers, carpenters, mechanics, and builders than any other mathematical law. Historical '& Fundamentals of High School Mathematics records show that the knowledge of this important relation is as old as civilization itself. EXERCISE 114 Problems based on the Pythagorean Theorem. 1. A rectangular schoolroom- floor is 32 feet long and 28 feet wide. What is the longest straight line that could be drawn upon the floor ? 2. How much walking is saved by cutting diago- nally across a rectangular plot of ground which is 25 rods wide and 42 rods long ? 3. A tree 100 feet high was broken off by a storm. The top struck the ground 40 feet from the foot of the tree, the broken end remaining on the stump. Find the height of the part standing, as- suming the ground to be level. Make a drawing. 4. What is the diagonal of a square whose sides are each 10 in. ? 5. Find the side of a square whose diagonal is 20 inches. 6. Two vessels start from the same place, one sail- ing due northwest at the rate of 12 miles per hour, and the other sailing due southwest at the rate of 16 miles per hour. How far apart are they at the end of 3 hours ? 7. The foot of a 36-foot ladder is 13 ft. 6 in. from the wall of a building against which the top is lean- ing. How high on the wall does the top reach ? 8. A rope stretched from the top of a 62-foot pole just reaches the ground 16 feet from the foot of the pole. Find the length of the rope. The Pythagorean Theorem *5V 9. The side of a square room is 21.5 feet. Find its diagonal correct to two decimals. 10. What is the perimeter of a square whose diag- onal is 12 inches ? 11. The side of a square is a. What represents its area ? its perimeter ? its diagonal ? 12. A rectangle is four times as long as it is wide. Find its diagonal if its area is 576 square inches. 13. Figure 115 is an equi- lateral triangle, and CD is perpendicular to AB. Find CD if each side of the triangle is 20 inches. Then find the area of triangle ABC 14. The area of a right triangle is 21 square inches. Its base is 6 inches, and hypotenuse. 15. The diagonal of a square is d. Show that s (side) is — -• 16. CD, the altitude of equilateral triangle ABC, is 16 inches. Find the sides of the triangle and its area. 17. How long an umbrella will lie flat down on the bottom of a trunk whose inside dimensions are 27 inches by 39 inches ? Find its altitude f\ 1 1 \ r1 Di \B \f k- ? — =» Fig. 116 258 Fundamentals of High School Mathematics 18. Can a circular wheel 8 feet in diameter be taken into a shop if the shop door is 4| feet wide and 6 1 feet high? 19. If A represents the area of a square, what will represent its perimeter ? its diagonal ? 20. The sides of a triangle are 12, 16, and 24 inches. Is it a right triangle ? Why ? 21. If you know two sides of a triangle, can you always find its area? Explain. 22. The hypotenuse of a right triangle is 10 feet, and one of its sides is 2 feet longer than the other. Find the length of the sides. 23. Find the area of a square whose diagonal is 12 inches longer than one of its sides. 24. Will an umbrella 30 inches long lie flat down in a suit case whose inside dimensions are 18 by 25 inches ? 25. A rectangle is 12 by 18 inches. How much must be added to its length to increase its diag- onal 4 inches ? 26. In a right triangle one side is one unit less than twice the other side. The hypotenuse is 17 units. What is the area of the triangle ? 27. One side of a right triangle is 3 times as long as the other. The hypotenuse contains 30 inches. Find the area of the triangle. 28. The dimensions of a certain rectangular black- board, and the longest line which can be drawn upon it, are represented in feet by three con- secutive even numbers. Find the dimensions of the blackboard. Solving Quadratic Equations with Two Unknowns 259 SOLVING QUADRATIC EQUATIONS WITH TWO UNKNOWNS Section 128. In the last chapter we solved quadratic equations in one unknown. Many quadratic equations, however, contain two, or more, unknowns. For example, consider the equation y = A which states that one number is equal to the square of another number. This equation contains two unknowns, x and j/, and is at the same time a second-degree or quad- ratic equation. A graph will help to show the relation between the variables or unknowns in this equation. Tabulating : Table 16 If x is o 1 2 3 4 5 -1 -2 -3 -4 then y is o 1 4 S 16 25 1 4 3 16 These values of x and y are plotted in Fig. 117. It is evident both from the graph and from the table that there \ _L t i i t -i i_ n r U ~i r^ ^ : t ~f -, "i /~ \ i : i^^i ±z _/\ ^r . Y Fig. 117. The line shows relationship between two variables, when one equals the square of the other. 260 Fundamentals of High School Mathematics is an indefinite number of sets of values of x and y wliich will satisfy the equation. Note that values of x are plotted along the horizontal axis, and values of y along the vertical axis. The graph of this equation may be thought of as answering the question which is suggested by the fol- lowing : What numbers are so related that one of them is equal to the square of the other ? EXERCISE 115 Graph each of the following quadratic equations : 1. y = x 2, + 4 X 5. x = y 2 -f- 6 y 2. y = x 2 — 4:X 6. x={y- 3)2 3. y=(x+2) 2 7. xy = 60 4. x = y 2 8. y = x 2 — x — 9. Illustrative exan lple. ( jraph the equa states that the sum of the squares of two numbers is 16. x 2 +y 2 = 16. Solving the equation for x gives 16 or Tabulating : x=±V16-i/2. Table 17 Ifyis 1 2 3' 4 5 -1 -2 -3 -4 -5 thenx is ±4. ±3.8 *3.4 ±2.6 * *3.8 ±34 ±2.6 * * If y is larger 1 becomes a nega ber. We call it han A, tive nu imagi then y 2 Tiber. nary. is gre; But w< iter ths i canno n 16 t ext and t -act tl he exp le squa ression re root under of a neg the r ;ative adica num- Solving Quadratic Equations with Two Unknowns 261 The graph of this equation is a circle whose radius is 4 units, as in Fig. 118. Y 04 n2.< 5,3 \5A ,2 3.8,1 - 5A-1 X Y Fig. 118 10. X 2 + y 2 = 25 11. f + x 1 = 40 GRAPHICAL SOLUTION OF A PAIR OF EQUATIONS Section 129. In solving equations there must always be as many equations as there are unknown quantities. The examples which you have just graphed were quadratic equations in two unknowns, or two variables. There are many sets of values of the unknowns which will satisfy any one of these equations, just as there were many sets of values which would satisfy a first-degree equation in two variables, such as x+j/ = 10. To obtain a single set of values, or a limited number of sets Rvalues which satisfy a quadratic equation in two unknowns, we must have two equations. (There must always be as many different equa- tions as there are unknowns.) 262 Fundamentals of High School Mathematics We shall now consider two equations : Illustrative example. (y 2 = 4x + 4. (1) I x + y = 2. (2) What set of values of x and y will satisfy both of these equations ? Let us first solve them graphically. Tabulating equation (1) 1/2 = 4* + 4. Table 18 If xis 2 3 8 -1 -2 then y is ±2. ±3.4 *4. ±6 impossible Plotting these' points gives the curve shown in Fig. 119. Y ^ \ / s A ■1 v^r 'u \ <i ^ "» s, J * s, ^ \ / s/ r f k f>\ \ V \ . \ V S, ^J s s ■^ T* > f ^ s fc * ^, % t S . Y Flo. 119. The x -distance and the ^-distance of the points of intersection of the two lines give the value of x and of y which satisfy the two equations. •X Solving Quadratic Equations with Two Unknowns 263 Plotting x -f y = 2 as in Chapter 11 gives the straight line. From Fig. 119 we see that the graphs intersect in two points, (0, 2) and (+8, — 6). Hence, there are two sets of values, and only two, which satisfy both equations : /* = • and (*=+ 8 \ y = 2 \y=-Q Show by checking that these values do satisfy both equations. The important step in solving equations graphically is to determine the intersection points of their graphs. If their graphs intersect in only one point, then there is only one set of values of the unknowns which will satisfy both equations. There will be as many solutions as there are intersection points of their graphs. EXERCISE 116 Solve graphically the following pairs of equations : 1. 2. ;tr 2 = 4jj/ lx 2 + y 2 = 16 2x+y = 12 3 ' [^+^ = 11.31 y 2 + 2=x \y = x 2 — 4:X ;r-3j/ = 4 * \2x-y = 5 ALGEBRAIC SOLUTION Section 130. The algebraic solution is much easier than the graphic solution. To illustrate, take the first example in the previous exercise : 264 Fundamentals of High School Mathematics Illustrative example. f* 2 = 4y, (i) \2x + y = 12. (2) From (2), y = 12-2x. (3) Substituting 12 — 2 x for y in (1) gives x 2 = 4(12 + 2*), (4) or X 2 = 4 8 _8x. (5) Solving by factoring, x 2 + 8* -48 = 0. • (6) (x+12)(x-4)=0. x = - 12 or + 4. If jc = — 12, thenz/ = 36. If x = 4, then y = 4. The pupil should check each pair of values. EXERCISE 117 Solve by the algebraic method each of the following pairs of equations. Check each. x + j/ = 6 5 * \j/=2x + l [2ab = - ! x*+2x=y 6 . -— - 2x+j>--= 12 [a-6 = 5 x+y = 2 \*^f = V 7 - W = -15 4. h+^ = 6 *£ = 5 1 2^r 2 — j/ 2 = — 9 9. Find two numbers whose difference is 9 and the sum of whose squares is 221. 10. The area of a rectangular field is 216 square rods, and its perimeter is 60 rods. What are its dimensions ? Solving Quadratic Equations with Two Unknowns 265 11. The sum of two numbers is \ 9 - and their product is |. Find the numbers. 12. The hypotenuse of a right triangle is 25 feet. Find the other two sides if you know that their sum is 35 feet. 13. A piece of wire 30 inches long is bent into the form of a right triangle whose hypotenuse is 13 inches. Find the other sides of the triangle. 14. The area and the perimeter of a rectangle are each 25. What are its dimensions ? 15. A photograph, 8 inches by 10 inches, is enlarged until it covers twice the original area, keeping the ratio of the length to the width unchanged. Find the sides of the enlarged photograph. 16. In placing telephone poles between two places, it was found that if the poles were set 10 feet farther apart than originally planned, 4 poles fewer per mile were needed. How far apart were the poles placed at first ? REVIEW EXERCISE 118 1. By substituting any value for x in x 2 — 1, 2jr, and x 2 + 1, show that the three numbers which result are sides of a right triangle. 2. If the sides of a right triangle are 6 inches and 8 inches, then the hypotenuse must be I inches. 3. How can you tell when a triangle is a right tri- angle without measuring its angle ? Is the triangle whose sides are 5, 12, and 13 a right triangle ? Why ? 266 Fundamentals of High School Mathematics 4. What is the area of an equilateral triangle each of whose sides is 30 inches? 5. How would you find the side of a square which had the same area as a circle with a radius of 12 inches ? 6. Write a formula for b if a, b, and c are the alti- tude, base, and hypotenuse of a right triangle ; similarly, a formula for a. 7. Express the hypotenuse of a right triangle whose altitude exceeds its base by 6 inches. ■ £ . \x + y = 12, 8. Solve the pair of equations: \ 2 ~ _ ^ q 9. When is it impossible to find the square root of a number ? A full and complete index will be supplied for the regular edition of this book. It has been decided to forego an index in the present Experimental Edition. LIBRARY 005 600 541 9 FW ■ ■ ■